## 1) Concise, abridged problem statement

We have a grammar over terminals `{a, b}` and nonterminals `1..N` (`N ≤ 30`).  
Each nonterminal `K` is defined as a sequence of terminals and/or nonterminals, where any referenced nonterminal is **< K** (so definitions form a DAG). Starting from the string consisting of just nonterminal `N`, repeatedly expand nonterminals by their definitions until only terminals remain; this is the **final string** (potentially extremely long).

Given a non-empty terminal string `S` (|S| ≤ 100), output the number of (possibly overlapping) occurrences of `S` in the final string, as a decimal integer (may be very large).

---

## 2) Detailed editorial (explaining the solution)

### Key difficulties
1. The expanded string of nonterminal `N` can be astronomically long, so we cannot build it.
2. We need to count occurrences of a pattern `S` in that implicit string, including overlaps.
3. The count can be huge ⇒ need big integers.

### KMP automaton for streaming pattern matching
We want to scan the final string left-to-right and count how many times pattern `S` ends at some position. This is classic for KMP:

- Compute prefix-function `pi` for `S` (length `m`).
- Build transition function `go[state][c]` where:
  - `state` = current matched prefix length (0..m)
  - `c` is terminal `a` or `b`
  - `go` returns the next matched prefix length after reading `c`.

This allows “streaming” matching: start at state 0; for every terminal char update state = go[state][c]; whenever state reaches `m`, we have matched an occurrence (ending here). (This implementation counts the occurrence when `state == m` after transition.)

### DP over nonterminals + KMP state
Each nonterminal expands into a sequence of elements, where each element is either:
- terminal `'a'` or `'b'`, or
- another nonterminal `< K`.

We need to know, for any nonterminal `K` and any starting KMP state `i`, what happens after scanning the expansion of `K`:

Define DP:
- `dp[K][i] = (count, state)`
  - `count` = number of occurrences of `S` found while scanning expansion of `K`, starting from KMP state `i`
  - `state` = final KMP state after scanning expansion of `K` starting from `i`

Why this works:
- Concatenation is composable: if a definition is `X Y`, then scanning `X` yields some count and end-state; scanning `Y` starting from that end-state yields additional count and final end-state. Counts add.

So to compute `dp[K][i]`, we simulate scanning K’s definition token-by-token:
- Maintain `cur_state` (starts at `i`) and `cnt` (big integer, starts 0).
- For each token:
  - If terminal:
    - `cur_state = go[cur_state][terminal]`
    - If `cur_state == m`, increment `cnt` by 1.
  - If nonterminal `t`:
    - Add `dp[t][cur_state].count` to `cnt`
    - Set `cur_state = dp[t][cur_state].state`

Because all referenced nonterminals are `< K`, we can compute `dp[0], dp[1], ..., dp[N-1]` in order (0-based in code).

Finally, the whole process starts from nonterminal `N` with KMP state 0:
- Answer is `dp[N][0].count`.

### Big integers
The occurrence count can be enormous (final string length can blow up exponentially with nesting), so the solution uses a `bigint` implementation:
- Base 1e9 storage, with addition and multiplication (Karatsuba) etc.
- Here we mainly rely on addition by 1 and adding other bigints.

### Complexity
Let:
- `N ≤ 30`
- `m = |S| ≤ 100`
- total number of tokens across definitions `≤ 500`

We compute `dp[K][i]` for each `K` and `i`:
- For each state `i` we iterate over all tokens in `K`’s definition.
- Total work: `O(m * total_tokens)` per nonterminal? More precisely:
  \[
  O\left((m+1)\cdot \sum_K L_K \right) \le O(101 \cdot 500) \approx 50k
  \]
plus big-int additions. This is easily within limits.

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>
using namespace std;

// Pretty-print a pair: "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read a pair from input
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read a whole vector: reads into each element in order
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) in >> x;
    return in;
};

// Print a vector elements separated by spaces
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) out << x << ' ';
    return out;
};

// Big integer base (each "digit" stores 0..1e9-1)
const int base = 1000000000;
const int base_digits = 9;

// Big integer structure supporting +, -, *, /, %, comparisons, etc.
struct bigint {
    vector<int> z; // digits in little-endian (z[0] is least significant)
    int sign;      // +1 or -1

    bigint() : sign(1) {}
    bigint(long long v) { *this = v; }
    bigint(const string& s) { read(s); }

    // Copy assignment
    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    // Assign from 64-bit integer
    void operator=(long long v) {
        sign = 1;
        if(v < 0) sign = -1, v = -v;
        z.clear();
        // Decompose v in base 1e9
        for(; v > 0; v /= base) z.push_back(v % base);
    }

    // Addition
    bigint operator+(const bigint& v) const {
        // If same sign, just add absolute values
        if(sign == v.sign) {
            bigint res = v; // start from v and add this->z into it
            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) res.z.push_back(0);
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) res.z[i] -= base;
            }
            return res;
        }
        // Different signs: a + (-b) = a - b
        return *this - (-v);
    }

    // Subtraction
    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            // If |this| >= |v|, do normal subtraction
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) res.z[i] += base;
                }
                res.trim(); // remove leading zeros
                return res;
            }
            // Otherwise result is negative: a - b = -(b - a)
            return -(v - *this);
        }
        // a - (-b) = a + b
        return *this + (-v);
    }

    // Multiply by int in-place
    void operator*=(int v) {
        if(v < 0) sign = -sign, v = -v;
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) z.push_back(0);
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
        }
        trim();
    }

    // Multiply by int (returns new bigint)
    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    // Division with remainder of two bigints: returns {quotient, remainder}
    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        // Normalize so that highest digit of b is large, improving quotient estimate
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        // Long division from most significant digit to least
        for(int i = (int)a.z.size() - 1; i >= 0; i--) {
            r *= base;      // shift remainder by one base digit
            r += a.z[i];    // bring down next digit

            // Estimate quotient digit using top digits
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = (int)(((long long)s1 * base + s2) / b.z.back());

            r -= b * d;     // subtract estimate * divisor
            while(r < 0) {  // if we overestimated, fix it
                r += b;
                --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm); // undo normalization on remainder
    }

    // Integer square root (not used in this problem)
    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);

        int n = (int)a.z.size();
        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) a.z.push_back(0);

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) { r = r1; break; }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = (int)(((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2));
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }
    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    // Divide by int in-place
    void operator/=(int v) {
        if(v < 0) sign = -sign, v = -v;
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    // Divide by int (returns new bigint)
    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    // Mod by int
    int operator%(int v) const {
        if(v < 0) v = -v;
        int m = 0;
        for(int i = (int)z.size() - 1; i >= 0; --i)
            m = (z[i] + m * (long long)base) % v;
        return m * sign;
    }

    // Compound ops
    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    // Comparisons
    bool operator<(const bigint& v) const {
        if(sign != v.sign) return sign < v.sign;
        if(z.size() != v.z.size())
            return (int)z.size() * sign < (int)v.z.size() * v.sign;
        for(int i = (int)z.size() - 1; i >= 0; i--)
            if(z[i] != v.z[i]) return z[i] * sign < v.z[i] * sign;
        return false;
    }
    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const { return !(*this < v) && !(v < *this); }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    // Remove leading zero digits and normalize sign
    void trim() {
        while(!z.empty() && z.back() == 0) z.pop_back();
        if(z.empty()) sign = 1;
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    // Unary minus
    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    // Absolute value
    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign; // +1
        return res;
    }

    // Convert to long long (unsafe if too big; not used here)
    long long longValue() const {
        long long res = 0;
        for(int i = (int)z.size() - 1; i >= 0; i--) res = res * base + z[i];
        return res * sign;
    }

    // gcd/lcm (not used here)
    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    // Read from string
    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        // parse optional +/- signs
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') sign = -sign;
            ++pos;
        }
        // parse digits in chunks of base_digits
        for(int i = (int)s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++)
                x = x * 10 + s[j] - '0';
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) stream << '-';
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i)
            stream << setw(base_digits) << setfill('0') << v.z[i];
        return stream;
    }

    // Convert digit vector between decimal chunk sizes (used for multiplication)
    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) p[i] = p[i - 1] * 10;

        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back((int)(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) res.pop_back();
        return res;
    }

    typedef vector<long long> vll;

    // Karatsuba multiplication on vectors (base 10^6 chunks)
    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = (int)a.size();
        vll res(n + n);

        // Small n: do O(n^2) multiplication
        if(n <= 32) {
            for(int i = 0; i < n; i++)
                for(int j = 0; j < n; j++)
                    res[i + j] += a[i] * b[j];
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        // (a1+a2)*(b1+b2)
        for(int i = 0; i < k; i++) a2[i] += a1[i];
        for(int i = 0; i < k; i++) b2[i] += b1[i];

        vll r = karatsubaMultiply(a2, b2);
        // r = (a1+a2)(b1+b2) - a1b1 - a2b2 = a1b2 + a2b1
        for(int i = 0; i < (int)a1b1.size(); i++) r[i] -= a1b1[i];
        for(int i = 0; i < (int)a2b2.size(); i++) r[i] -= a2b2[i];

        // Assemble result
        for(int i = 0; i < (int)r.size(); i++) res[i + k] += r[i];
        for(int i = 0; i < (int)a1b1.size(); i++) res[i] += a1b1[i];
        for(int i = 0; i < (int)a2b2.size(); i++) res[i + n] += a2b2[i];
        return res;
    }

    // Bigint * bigint using Karatsuba (overkill for this task, but fine)
    bigint operator*(const bigint& v) const {
        // Convert from base 1e9 digits to base 1e6 digits for safe multiplication
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());

        // Pad to equal length
        while(a.size() < b.size()) a.push_back(0);
        while(b.size() < a.size()) b.push_back(0);

        // Pad length to power of two (required by this Karatsuba implementation)
        while(a.size() & (a.size() - 1)) {
            a.push_back(0);
            b.push_back(0);
        }

        // Multiply coefficient vectors
        vll c = karatsubaMultiply(a, b);

        bigint res;
        res.sign = sign * v.sign;

        // Carry handling in base 1e6
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }

        // Convert back to base 1e9
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

// Problem input data
int n;
string s;
vector<vector<int>> defs; // defs[k] is definition of nonterminal k (0-based):
                          // terminals encoded as -1 ('a') and -2 ('b'),
                          // nonterminals stored as index (0..k-1).

// Read input and build defs
void read() {
    cin >> n;     // number of nonterminals
    cin >> s;     // pattern string S
    defs.assign(n, {});
    for(auto& def: defs) {
        int sz;
        cin >> sz;        // length of definition
        def.resize(sz);
        for(int j = 0; j < sz; j++) {
            string tok;
            cin >> tok;
            if(tok == "a") {
                def[j] = -1;
            } else if(tok == "b") {
                def[j] = -2;
            } else {
                def[j] = stoi(tok) - 1; // convert 1-based nonterminal to 0-based
            }
        }
    }
}

void solve() {
    int m = (int)s.size(); // pattern length

    // Compute KMP prefix function pi for pattern s
    vector<int> pi(m, 0);
    for(int i = 1; i < m; i++) {
        int j = pi[i - 1];
        while(j > 0 && s[j] != s[i]) j = pi[j - 1];
        if(s[j] == s[i]) j++;
        pi[i] = j;
    }

    // Build automaton transitions go[state][c] for c in {0:'a', 1:'b'}
    // state ranges 0..m (inclusive); m means full match just happened.
    vector<array<int, 2>> go(m + 1);
    for(int i = 0; i <= m; i++) {
        for(int c = 0; c < 2; c++) {
            char ch = "ab"[c];
            if(i < m && s[i] == ch) {
                // Next character extends current matched prefix
                go[i][c] = i + 1;
            } else if(i == 0) {
                // From state 0, mismatch stays in 0
                go[i][c] = 0;
            } else {
                // Fallback using prefix function
                go[i][c] = go[pi[i - 1]][c];
            }
        }
    }

    // dp[k][i] = {count_occurrences, ending_kmp_state}
    // for expansion of nonterminal k starting from KMP state i.
    vector<vector<pair<bigint, int>>> dp(n, vector<pair<bigint, int>>(m + 1));

    // Process nonterminals in increasing order (DAG topological order)
    for(int k = 0; k < n; k++) {
        // For every possible starting KMP state
        for(int i = 0; i <= m; i++) {
            int cur = i;     // current KMP state while scanning expansion
            bigint cnt = 0;  // total matches found so far

            // Scan the definition sequence left-to-right
            for(int elem: defs[k]) {
                if(elem < 0) {
                    // Terminal character
                    int c = (elem == -1) ? 0 : 1; // -1 -> 'a', -2 -> 'b'
                    cur = go[cur][c];             // advance KMP automaton
                    if(cur == m) {
                        // Full match found ending here
                        cnt += 1;
                    }
                } else {
                    // Nonterminal reference: compose precomputed result
                    cnt += dp[elem][cur].first;   // add matches inside elem expansion
                    cur = dp[elem][cur].second;   // update state after elem expansion
                }
            }

            dp[k][i] = {cnt, cur}; // store result for this start state
        }
    }

    // Start from nonterminal N (index n-1) with empty prefix matched (state 0)
    cout << dp[n - 1][0].first << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;  // only one test in this problem
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }
    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys

def build_kmp_automaton(pattern: str):
    """
    Build KMP prefix-function and automaton transitions for alphabet {a,b}.
    Returns: go, where go[state][c] gives next state
             state in [0..m], c in {0,1} for 'a','b'
    """
    m = len(pattern)

    # Prefix-function pi[i] = length of longest proper prefix of pattern
    # which is also a suffix of pattern[:i+1]
    pi = [0] * m
    for i in range(1, m):
        j = pi[i - 1]
        while j > 0 and pattern[j] != pattern[i]:
            j = pi[j - 1]
        if pattern[j] == pattern[i]:
            j += 1
        pi[i] = j

    # Automaton transitions
    go = [[0, 0] for _ in range(m + 1)]
    for state in range(m + 1):
        for c in range(2):
            ch = 'a' if c == 0 else 'b'
            if state < m and pattern[state] == ch:
                # Successful extension
                go[state][c] = state + 1
            elif state == 0:
                # Can't fallback further
                go[state][c] = 0
            else:
                # Fallback and reuse transition
                go[state][c] = go[pi[state - 1]][c]

    return go

def solve():
    data = sys.stdin.read().strip().split()
    it = iter(data)

    n = int(next(it))        # number of nonterminals
    s = next(it)             # pattern S
    m = len(s)

    # Read grammar definitions.
    # We store each definition as a list of tokens:
    #   -1 = 'a', -2 = 'b', >=0 = referenced nonterminal index
    defs = []
    for _ in range(n):
        L = int(next(it))
        cur = []
        for _ in range(L):
            tok = next(it)
            if tok == 'a':
                cur.append(-1)
            elif tok == 'b':
                cur.append(-2)
            else:
                cur.append(int(tok) - 1)  # convert 1-based to 0-based
        defs.append(cur)

    # KMP automaton
    go = build_kmp_automaton(s)

    # dp[k][state] = (count, end_state)
    # In Python, big integers are built-in, so count is just int.
    dp_count = [[0] * (m + 1) for _ in range(n)]
    dp_state = [[0] * (m + 1) for _ in range(n)]

    # Compute DP in increasing order of nonterminal index (DAG order)
    for k in range(n):
        for start_state in range(m + 1):
            cur_state = start_state
            cnt = 0

            # Stream through definition tokens, composing results
            for elem in defs[k]:
                if elem < 0:
                    # Terminal
                    c = 0 if elem == -1 else 1
                    cur_state = go[cur_state][c]
                    if cur_state == m:
                        cnt += 1
                else:
                    # Nonterminal: add its internal matches and jump end state
                    cnt += dp_count[elem][cur_state]
                    cur_state = dp_state[elem][cur_state]

            dp_count[k][start_state] = cnt
            dp_state[k][start_state] = cur_state

    # Answer: start from nonterminal N with KMP state 0
    print(dp_count[n - 1][0])

if __name__ == "__main__":
    solve()
```

---

## 5) Compressed editorial

- Build KMP prefix-function for pattern `S` and automaton `go[state][a/b]` for states `0..m`.
- DP over nonterminals in increasing order (since each `K` only uses `<K`):
  - `dp[K][i] = (cnt, st)` where scanning expansion of `K` starting from KMP state `i` produces `cnt` matches and ends in state `st`.
  - Compute by simulating the definition tokens:
    - For terminal: `st = go[st][c]`, if `st==m` then `cnt++`.
    - For nonterminal `t`: `cnt += dp[t][st].cnt`, `st = dp[t][st].st`.
- Output `dp[N][0].cnt`. Use big integers (C++ bigint; Python int).