<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

284. Grammar
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Consider an alphabet consisting of terminal characters 'a' and 'b' and non-terminal characters 1,2,...,N. Each non-terminal character K has a description, which is a string of terminal and non-terminal characters; the non-terminal characters in the description are less than K. The description may be empty.

You take the string containing single non-terminal character N and replace all non-terminal characters in it by their definitions until we obtain a string containing only terminal characters, which is called the final string.

The task is: given a nonempty string S of terminal characters, determine the number of its occurences in the final string.

Input
The first line of the input contains single integer N (1≤ N≤ 30). The second line contains string S containing at most 100 characters. The rest of the input contains the non-terminal characters' descriptions. The K+2-nd line (1≤ K≤ N) contains the description of K-th character. It starts with a non-negative integer LK, followed by LK characters of the alphabet separated by spaces. The sum of all LK does not exceed 500.

Output
The only line of the output must contain the answer without leading zeroes.

Example(s)
sample input
sample output
2
abb
2 a b 
3 a 1 b
1



Novosibirsk SU Contest #2, by Novosibirsk Team #1

<|response|>
## 1) Abridged problem statement (concise)

We have terminals `{a, b}` and nonterminals `1..N` (`N ≤ 30`).  
Each nonterminal `K` is defined as a sequence of terminals and/or nonterminals, and every referenced nonterminal is `< K` (so the dependency graph is a DAG). Starting from the string `"N"`, repeatedly replace nonterminals by their definitions until only terminals remain; this yields the (possibly enormous) **final string**.

Given a non-empty terminal pattern `S` (`|S| ≤ 100`), compute how many times `S` occurs (overlaps allowed) in the final string. The answer can be extremely large, so print it as a decimal integer.

---

## 2) Key observations

1. **We cannot build the final string**: expansions may grow exponentially.
2. We need to count occurrences of a pattern in a huge implicit string ⇒ use **streaming pattern matching**.
3. With KMP, we can represent the scanning process by a small **automaton state**:
   - state = length of matched prefix of `S` (0..m), where `m = |S|`.
   - reading `'a'` or `'b'` updates the state via precomputed transitions.
   - whenever we reach state `m`, we found an occurrence ending at this position.
4. Since each nonterminal `K` only references `< K`, we can compute DP **in increasing order of K**.
5. For each nonterminal expansion, what matters for pattern counting is:
   - how many matches occur inside it
   - what KMP state we end in, given the starting KMP state

This suggests a composable DP.

---

## 3) Full solution approach

### Step A: Build KMP automaton for pattern `S`

Let `S` have length `m`. Compute the prefix-function `pi[]`.  
Then build automaton transitions:

`go[state][c]` for `state ∈ [0..m]` and `c ∈ {0,1}` corresponding to `'a'` / `'b'`.

This allows scanning any terminal stream without backtracking:
- `state = go[state][c]`
- if `state == m`, count one occurrence (overlaps naturally handled by KMP).

### Step B: DP over nonterminals and KMP start state

Encode each definition token:
- terminal `'a'` as `-1`, terminal `'b'` as `-2`
- nonterminal `t` as integer index `t-1` (0-based)

Define DP:

For each nonterminal `k` (0-based) and each KMP start state `i` (0..m):
- `dpCount[k][i]` = number of occurrences found while scanning expansion of `k` starting from state `i`
- `dpState[k][i]` = resulting KMP state after scanning expansion of `k` starting from state `i`

To compute `(dpCount[k][i], dpState[k][i])`:
- initialize `curState = i`, `cnt = 0`
- scan tokens in `defs[k]` left-to-right:
  - if terminal:
    - `curState = go[curState][a/b]`
    - if `curState == m`, `cnt++`
  - if nonterminal `t`:
    - `cnt += dpCount[t][curState]`
    - `curState = dpState[t][curState]`

Because all referenced nonterminals are `< k`, their DP is already computed.

### Step C: Output

The final process starts at nonterminal `N` (index `N-1`) with KMP state `0`.

Answer = `dpCount[N-1][0]`.

### Big integers
The count can be huge.  
- In **Python**, `int` is arbitrary precision.
- In **C++**, use `boost::multiprecision::cpp_int` (much simpler than writing a bigint).

### Complexity
Let total tokens across all definitions be `T ≤ 500`, and `m ≤ 100`.
We do `O((m+1) * T)` transitions per nonterminal set overall (effectively `O((m+1)*T)` because we scan each rule for each state), well within limits.

---

## 4) C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

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

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

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

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            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;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            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();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    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);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        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());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            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 = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                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);
    }

    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 = 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 = ((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; }

    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();
    }

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

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

    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; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = 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; }

    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]); }

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

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    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;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = 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;
    }

    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;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        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);

        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);
        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];
        }

        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 operator*(const bigint& v) const {
        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());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        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);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int n;
string s;
vector<vector<int>> defs;

void read() {
    cin >> n;
    cin >> s;
    defs.assign(n, {});
    for(auto& def: defs) {
        int sz;
        cin >> sz;
        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;
            }
        }
    }
}

void solve() {
    // We should first note that the definitions form a DAG, and unfortunately,
    // this might mean the final length is quite large and we would have to use
    // big integers. Afterwards, this problem is a clear example of DP. We can
    // perform KMP over the given string so that we can easily precompute an
    // array nxt[position][char] - the longest prefix of s that appears as a
    // suffix of s[:position] + char, for char being one of {a, b}. Then we can
    // try maintaining the following state:
    //
    //     dp[K][prefix i of s] = (number of occurrences of s in K,
    //                             longest prefix of s that is a suffix of K)
    //
    // Then the answer is dp[N][K]. We can compute this recursively as the
    // definitions graph is a DAG. The key implementation detail is to maintain
    // the result of the DP as a pair that also contains the resulting prefix.

    int m = s.size();

    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;
    }

    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) {
                go[i][c] = i + 1;
            } else if(i == 0) {
                go[i][c] = 0;
            } else {
                go[i][c] = go[pi[i - 1]][c];
            }
        }
    }

    vector<vector<pair<bigint, int>>> dp(n, vector<pair<bigint, int>>(m + 1));

    for(int k = 0; k < n; k++) {
        for(int i = 0; i <= m; i++) {
            int cur = i;
            bigint cnt = 0;
            for(int elem: defs[k]) {
                if(elem < 0) {
                    int c = (elem == -1) ? 0 : 1;
                    cur = go[cur][c];
                    if(cur == m) {
                        cnt += 1;
                    }
                } else {
                    cnt += dp[elem][cur].first;
                    cur = dp[elem][cur].second;
                }
            }
            dp[k][i] = {cnt, cur};
        }
    }

    cout << dp[n - 1][0].first << "\n";
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 5) Python implementation (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} representing 'a','b'.
    """
    m = len(pattern)

    # Prefix function pi[i] = length of longest proper prefix of pattern
    # that 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:
                go[state][c] = state + 1
            elif state == 0:
                go[state][c] = 0
            else:
                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))
    S = next(it)
    m = len(S)

    # Read definitions and encode tokens:
    # -1 = 'a', -2 = 'b', >=0 = nonterminal index (0-based)
    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)
        defs.append(cur)

    # KMP automaton transitions
    go = build_kmp_automaton(S)

    # DP tables
    # Python ints are arbitrary precision, perfect for this problem.
    dpCount = [[0] * (m + 1) for _ in range(N)]
    dpState = [[0] * (m + 1) for _ in range(N)]

    # Compute dp in increasing nonterminal order
    for k in range(N):
        for start in range(m + 1):
            cur_state = start
            cnt = 0

            for token in defs[k]:
                if token < 0:
                    # Terminal
                    c = 0 if token == -1 else 1
                    cur_state = go[cur_state][c]
                    if cur_state == m:
                        cnt += 1
                else:
                    # Nonterminal reference: compose results
                    cnt += dpCount[token][cur_state]
                    cur_state = dpState[token][cur_state]

            dpCount[k][start] = cnt
            dpState[k][start] = cur_state

    # Answer: expand N starting with KMP state 0
    print(dpCount[N - 1][0])

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

Both implementations follow the same idea: treat expansions as streams, use the KMP automaton state as the only “context”, and do DP composition along the grammar DAG.