<|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.

201. Non Absorbing DFA
time limit per test: 0.25 sec.
memory limit per test: 65536 KB
input: standard
output: standard



In the theory of compilers and languages finite state machines, also known as finite automata are widely used. Deterministic finite automation (DFA) is an ordered set <Σ, U, s, T, φ> where Σ is the finite set called input alphabet, U is the finite set of states, s from U is the initial state, T subset of U is the set of terminal states and φ : U × Σ → U is the transition function.

The input of the automation is the string α over Σ. Initially the automation is in state s. Each step it reads the first character c of the input string and changes its state to φ(u, c) where u is the current state. After that the first character of the input string is removed and the step repeats. If when its input string is empty the automation is in the terminal state, it is said that it accepts the initial string α, in the other case it rejects it.

In some cases to simplify the automation the concept of nonabsorbing edges is introduced. That is, in addition to φ the function χ : U × Σ → {0, 1} is introduced and when making a transition from some state u with some character c, the leading character is removed from the input string only if χ(u, c) = 0. If χ(u, c) = 1, the input string is kept intact and next transition is performed with the new state and the same character.

It is said that such automation accepts some string α if after a number of steps it transits to the terminal state and the input string becomes empty.

Your task is given the DFA with nonabsorbing edges to compute the number of strings of the given length N that it accepts.

Input

The first line of the input file contains Σ — a subset of the English alphabet, several different small letters. Next line contains K = | U | — the number of states of the automation (1 ≤ K ≤ 1000). Let states be numbered from 1 to K. Next line contains S (1 ≤ S ≤ K) — the initial state, followed by L = | T | — the number of terminal states and then L different integer numbers ranging from 1 to K — the numbers of terminal states.

Next K lines contain | Σ | integer numbers each and define φ. Next K lines define χ in a similar way. The last line of the input file contains N (1 ≤ N ≤ 60).

Output

Output the only number — the number of different strings of length N over Σ that the given DFA accepts.

In the given example the two strings accepted by the automation are ``aaa'' and ``abb''.

Sample test(s)

Input
ab
2
1 1 2
2 1
1 2
0 1
0 0
3

Output
2

Author:	Andrew Stankevich
Resource:	Petrozavodsk Summer Trainings 2003
Date:	2003-08-23

<|response|>
1. Abridged Problem Statement  
We have a deterministic finite automaton (DFA) extended with “non-absorbing” edges. Formally:  
• Alphabet Σ of size σ.  
• States U = {1…K}, initial state S, terminal states T.  
• Transition φ: U×Σ→U, and absorption flag χ: U×Σ→{0,1}.  
  – If χ(u,c)=0 the input character c is consumed; if χ(u,c)=1 it is not consumed.  
A string is accepted if, after a finite number of steps, the automaton reaches a terminal state and the input is empty. Count how many distinct strings of length exactly N over Σ are accepted. Constraints: K≤1000, N≤60, σ≤26.

2. Key Observations  
– A non-absorbing edge may move between states without consuming the current character, possibly forming loops that never consume it. Any string that attempts to use such a loop can never finish consuming N characters and accept.  
– We can “collapse” each (state u, letter c) into one effective absorbing move: follow χ=1 edges repeatedly until you hit a state that consumes c (χ=0) or detect a cycle of χ=1 edges (infinite loop).  
– After collapsing, every letter consumption becomes a single transition: from u reading c you end up in some state v=φ(a,c), where a is the state that actually consumes c.  
– We then do a standard DP over the number of consumed characters: dp[i][u] = number of length-i prefixes that lead to state u.  

3. Full Solution Approach  
Step A. Read input: Σ, K, S, terminal states T, tables φ and χ, and N. Convert states to 0-based.  
Step B. Precompute effective transitions:  
  Create an array trans[K][σ], initially marked UNVISITED.  
  Define a DFS procedure dfs(u,c):  
    if χ[u][c]==0 then trans[u][c]=u (consumes here) and return.  
    if trans[u][c] is being visited (mark VISITING) then we found a χ=1 cycle ⇒ set trans[u][c] = -1 (infinite loop) and return.  
    otherwise mark trans[u][c]=VISITING, let v=φ[u][c], call dfs(v,c), then set trans[u][c] = trans[v][c].  
  Run dfs(u,c) for all u in [0..K−1], c in [0..σ−1].  
Step C. Dynamic Programming:  
  Let dp be a (N+1)×K table of big integers, all zero. Set dp[0][S]=1.  
  For i from 0 to N−1, for each state u, if dp[i][u]>0 then for each letter c:  
    a = trans[u][c]; if a<0 skip (infinite loop).  
    v = φ[a][c];  
    dp[i+1][v] += dp[i][u].  
Step D. The answer is sum(dp[N][t]) over all terminal states t. Print it.  

Time complexity: O(K·σ + N·K·σ). N≤60, K≤1000, σ≤26 is fine.  

4. C++ Implementation
```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 {
        bigint res = *this;
        res += v;
        return res;
    }

    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);
    }

    bigint& operator+=(const bigint& v) {
        if(sign == v.sign) {
            int carry = 0;
            if(z.size() < v.z.size()) {
                z.resize(v.z.size(), 0);
            }
            for(size_t i = 0; i < max(z.size(), v.z.size()) || carry; ++i) {
                if(i == z.size()) {
                    z.push_back(0);
                }
                long long cur = z[i] + carry + (i < v.z.size() ? v.z[i] : 0LL);
                carry = cur >= base;
                if(carry) {
                    cur -= base;
                }
                z[i] = (int)cur;
            }
            return *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);
        }
        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;
    }

    bigint& operator-=(const bigint& v) {
        *this = *this - v;
        return *this;
    }
    bigint& operator*=(const bigint& v) {
        *this = *this * v;
        return *this;
    }
    bigint& operator/=(const bigint& v) {
        *this = *this / v;
        return *this;
    }

    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;
    }

    void reset() {
        z.assign(1, 0);
        sign = 1;
    }
};

string alphabet;
int alphabet_size;
int k;
int initial_state;
vector<int> terminal_states;
vector<vector<int>> phi;
vector<vector<int>> chi;
int n;

void read() {
    cin >> alphabet;
    alphabet_size = alphabet.size();

    cin >> k;
    cin >> initial_state;
    initial_state--;

    int num_terminal;
    cin >> num_terminal;

    terminal_states.resize(num_terminal);
    for(int i = 0; i < num_terminal; i++) {
        cin >> terminal_states[i];
        terminal_states[i]--;
    }

    phi.assign(k, vector<int>(alphabet_size));
    for(int i = 0; i < k; i++) {
        for(int j = 0; j < alphabet_size; j++) {
            cin >> phi[i][j];
            phi[i][j]--;
        }
    }

    chi.assign(k, vector<int>(alphabet_size));
    for(int i = 0; i < k; i++) {
        for(int j = 0; j < alphabet_size; j++) {
            cin >> chi[i][j];
        }
    }

    cin >> n;
}

void solve() {
    // - A non-absorbing edge (chi = 1) keeps the current character and moves
    // on,
    //   so for every (state, char) we resolve the chain of non-absorbing edges
    //   to the first state where the character is actually consumed.
    //   trans[s][c] stores that absorbing state, with -1 marking a
    //   non-absorbing cycle (the char can never be consumed from there) and -2
    //   meaning not yet computed. A DFS along phi[.][c] fills trans, painting
    //   -1 while in progress to catch cycles.
    //
    // - dp[i][state] counts length-i prefixes that leave the automaton in
    //   state. From dp[length][cur], each character c first slides to its
    //   absorbing state, then consumes one symbol via phi, contributing to
    //   dp[length + 1][phi[absorbing][c]]; cycle edges (-1) are skipped.
    //
    // - The answer sums dp[n][.] over the terminal states. Counts reach
    //   alphabet_size^n, so they are accumulated as big integers.

    vector<vector<int>> trans(k, vector<int>(alphabet_size, -2));

    function<void(int, int)> dfs = [&](int state, int ch) {
        if(chi[state][ch] == 0) {
            trans[state][ch] = state;
            return;
        }

        if(trans[state][ch] != -2) {
            return;
        }

        trans[state][ch] = -1;

        int next_state = phi[state][ch];
        dfs(next_state, ch);
        trans[state][ch] = trans[next_state][ch];
    };

    for(int state = 0; state < k; state++) {
        for(int ch = 0; ch < alphabet_size; ch++) {
            if(trans[state][ch] == -2) {
                dfs(state, ch);
            }
        }
    }

    vector<vector<int>> next_state(k, vector<int>(alphabet_size, -1));
    for(int state = 0; state < k; state++) {
        for(int ch = 0; ch < alphabet_size; ch++) {
            if(trans[state][ch] >= 0) {
                next_state[state][ch] = phi[trans[state][ch]][ch];
            }
        }
    }

    vector<bigint> cur(k), nxt(k);
    vector<char> in_cur(k, 0), in_nxt(k, 0);
    vector<int> list_cur, list_nxt;

    cur[initial_state] = bigint(1);
    in_cur[initial_state] = 1;
    list_cur.push_back(initial_state);

    for(int length = 0; length < n; length++) {
        for(int current_state: list_cur) {
            const bigint& ways = cur[current_state];
            if(ways.isZero()) {
                continue;
            }

            for(int ch = 0; ch < alphabet_size; ch++) {
                int to = next_state[current_state][ch];
                if(to < 0) {
                    continue;
                }
                if(!in_nxt[to]) {
                    in_nxt[to] = 1;
                    nxt[to].reset();
                    list_nxt.push_back(to);
                }
                nxt[to] += ways;
            }

            cur[current_state].reset();
            in_cur[current_state] = 0;
        }

        list_cur.swap(list_nxt);
        list_nxt.clear();
        swap(cur, nxt);
        swap(in_cur, in_nxt);
    }

    bigint result;
    for(int terminal_state: terminal_states) {
        result += cur[terminal_state];
    }

    cout << result << '\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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

5. Python Implementation with Detailed Comments  
```python
import sys
sys.setrecursionlimit(10000)

def main():
    # 1. Read input
    alphabet = sys.stdin.readline().strip()
    sigma = len(alphabet)

    K = int(sys.stdin.readline())
    parts = list(map(int, sys.stdin.readline().split()))
    S = parts[0] - 1             # zero-based
    L = parts[1]
    terminals = [x-1 for x in parts[2:]]

    # φ table
    phi = [None]*K
    for u in range(K):
        row = list(map(int, sys.stdin.readline().split()))
        phi[u] = [x-1 for x in row]

    # χ table
    chi = [None]*K
    for u in range(K):
        chi[u] = list(map(int, sys.stdin.readline().split()))

    N = int(sys.stdin.readline())

    # 2. Precompute effective transitions
    # trans[u][c] = state where c is consumed, or -1 if infinite loop
    UNVIS, VISITING = None, -1
    trans = [[UNVIS]*sigma for _ in range(K)]

    def dfs(u, c):
        if chi[u][c] == 0:
            # consumes here
            trans[u][c] = u
            return
        if trans[u][c] == VISITING:
            # found a non-consuming cycle
            trans[u][c] = -1
            return
        if trans[u][c] is not UNVIS:
            # already resolved
            return
        # mark visiting
        trans[u][c] = VISITING
        v = phi[u][c]
        dfs(v, c)
        trans[u][c] = trans[v][c]

    for u in range(K):
        for c in range(sigma):
            if trans[u][c] is UNVIS:
                dfs(u, c)

    # 3. DP: dp[i][u] = number of ways to reach u after i consumed chars
    dp = [ [0]*K for _ in range(N+1) ]
    dp[0][S] = 1

    for i in range(N):
        for u in range(K):
            ways = dp[i][u]
            if ways == 0:
                continue
            for c in range(sigma):
                a = trans[u][c]
                if a < 0:
                    continue   # infinite loop
                v = phi[a][c]
                dp[i+1][v] += ways

    # 4. Sum over terminal states
    result = sum(dp[N][t] for t in terminals)
    print(result)

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

Explanation of the Key Steps:  
- We collapse any chain of non-consuming edges into one effective “where will this letter be consumed?” transition, marking those chains that never consume (pure loops) as invalid.  
- With these effective transitions, each input character simply advances the DP by one consumed character, just like in a normal DFA, and we never worry again about χ flags.  
- A final sum over terminal states gives the total number of accepted strings of length N.