<|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++ implementation (detailed comments)

```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>

using namespace std;
using boost::multiprecision::cpp_int;

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

    int N;
    cin >> N;

    string S;
    cin >> S;
    int m = (int)S.size();

    // Read definitions.
    // Encode tokens:
    //  -1 => 'a', -2 => 'b', >=0 => nonterminal index (0-based)
    vector<vector<int>> defs(N);
    for (int k = 0; k < N; k++) {
        int L;
        cin >> L;
        defs[k].resize(L);
        for (int i = 0; i < L; i++) {
            string tok;
            cin >> tok;
            if (tok == "a") defs[k][i] = -1;
            else if (tok == "b") defs[k][i] = -2;
            else defs[k][i] = stoi(tok) - 1; // 1-based -> 0-based
        }
    }

    // ---- Build KMP prefix function ----
    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 from 0..m (m means "we have matched full pattern length",
    // but we still allow transitions from it for convenience).
    vector<array<int, 2>> go(m + 1);
    for (int state = 0; state <= m; state++) {
        for (int c = 0; c < 2; c++) {
            char ch = (c == 0 ? 'a' : 'b');
            if (state < m && S[state] == ch) {
                go[state][c] = state + 1;            // extend match
            } else if (state == 0) {
                go[state][c] = 0;                    // cannot fallback
            } else {
                go[state][c] = go[pi[state - 1]][c]; // fallback via pi
            }
        }
    }

    // ---- DP arrays ----
    // dpCount[k][i] = how many matches occur when expanding nonterminal k,
    // starting KMP state i.
    // dpState[k][i] = ending KMP state after expanding nonterminal k,
    // starting KMP state i.
    vector<vector<cpp_int>> dpCount(N, vector<cpp_int>(m + 1, 0));
    vector<vector<int>> dpState(N, vector<int>(m + 1, 0));

    // Process nonterminals in increasing order (DAG order guaranteed by input).
    for (int k = 0; k < N; k++) {
        for (int start = 0; start <= m; start++) {
            int curState = start;
            cpp_int cnt = 0;

            // Scan the definition tokens left-to-right and compose effects.
            for (int token : defs[k]) {
                if (token < 0) {
                    // Terminal
                    int c = (token == -1 ? 0 : 1);
                    curState = go[curState][c];
                    if (curState == m) {
                        // A full occurrence ends here
                        cnt += 1;
                    }
                } else {
                    // Nonterminal reference: add its internal matches
                    // and jump state accordingly.
                    cnt += dpCount[token][curState];
                    curState = dpState[token][curState];
                }
            }

            dpCount[k][start] = cnt;
            dpState[k][start] = curState;
        }
    }

    // Start from nonterminal N (index N-1) with KMP state 0.
    cout << dpCount[N - 1][0] << "\n";
    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.