Fibonacci strings are defined as follows: 

  * f1 = «a» 
  * f2 = «b» 
  * fn = fn - 1 fn - 2, n > 2



Thus, the first five Fibonacci strings are: "a", "b", "ba", "bab", "babba".

You are given a Fibonacci string and m strings si. For each string si, find the number of times it occurs in the given Fibonacci string as a substring.

Input

The first line contains two space-separated integers k and m — the number of a Fibonacci string and the number of queries, correspondingly.

Next m lines contain strings si that correspond to the queries. It is guaranteed that strings si aren't empty and consist only of characters "a" and "b".

The input limitations for getting 30 points are: 

  * 1 ≤ k ≤ 3000
  * 1 ≤ m ≤ 3000
  * The total length of strings si doesn't exceed 3000



The input limitations for getting 100 points are: 

  * 1 ≤ k ≤ 1018
  * 1 ≤ m ≤ 104
  * The total length of strings si doesn't exceed 105



Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

For each string si print the number of times it occurs in the given Fibonacci string as a substring. Since the numbers can be large enough, print them modulo 1000000007 (109 + 7). Print the answers for the strings in the order in which they are given in the input.

Examples

Input

6 5
a
b
ab
ba
aba


Output

3
5
3
3
1
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
struct matrix {
  int M[4][4];
  matrix() { memset(M, 0, sizeof(M)); }
};
matrix one, stand;
matrix operator*(matrix M1, matrix M2) {
  matrix M3;
  for (int i = 0; i < 4; ++i)
    for (int j = 0; j < 4; ++j)
      for (int k = 0; k < 4; ++k)
        M3.M[i][j] = (M3.M[i][j] + 1LL * M1.M[i][k] * M2.M[k][j]) % MOD;
  return M3;
}
matrix power(matrix X, long long ex) {
  if (ex == 0) return one;
  if (ex & 1) return X * power(X, ex - 1);
  return power(X * X, ex >> 1);
}
long long N, M;
int Q[100002], C[4];
char A[100002];
int get_times(int size, string S, int lim1, int lim2) {
  int qnow = 0;
  for (int i = 2; i <= size; ++i) {
    while (qnow != 0 && A[i] != A[qnow + 1]) qnow = Q[qnow];
    if (A[i] == A[qnow + 1]) ++qnow;
    Q[i] = qnow;
  }
  int matchnow = 0, totmatches = 0;
  for (int i = 0; i < int(S.size()); ++i) {
    while (matchnow != 0 && S[i] != A[matchnow + 1]) matchnow = Q[matchnow];
    if (S[i] == A[matchnow + 1]) ++matchnow;
    if (matchnow == size) {
      if (i - size + 1 >= lim1 && i - size + 1 <= lim2) ++totmatches;
      matchnow = Q[matchnow];
    }
  }
  return totmatches;
}
int main() {
  for (int i = 0; i < 4; ++i) one.M[i][i] = 1;
  stand.M[0][0] = stand.M[1][0] = stand.M[2][0] = stand.M[0][1] =
      stand.M[3][2] = stand.M[2][3] = 1;
  cin >> N >> M;
  for (int i = 1; i <= M; ++i) {
    cin >> (A + 1);
    int size = 0;
    for (int j = 1; A[j] != '\0'; ++j) ++size;
    int fib1 = 1, fib2 = 1, numnow = 0;
    while (fib1 < size) {
      int auxfib = fib2;
      fib2 = fib1 + fib2;
      fib1 = auxfib;
      ++numnow;
    }
    string s1, s2;
    s1 = "a";
    s2 = "b";
    for (int j = 1; j <= numnow; ++j) {
      string saux = s2;
      s2 += s1;
      s1 = saux;
    }
    if (N - (numnow + 1) <= 1) {
      if (N - (numnow + 1) < 0) {
        cout << 0 << '\n';
        continue;
      }
      while (N - (numnow + 1) > 0) {
        string saux = s2;
        s2 += s1;
        s1 = saux;
        ++numnow;
      }
      cout << get_times(size, s1, 0, s1.size() - 1) << '\n';
      continue;
    }
    int times_a = get_times(size, s1, 0, s1.size() - 1);
    int times_b = get_times(size, s2, 0, s2.size() - 1);
    int times_ab =
        get_times(size, s1 + s2, s1.size() - size + 1, s1.size() - 1);
    int times_ba =
        get_times(size, s2 + s1, s2.size() - size + 1, s2.size() - 1);
    int times_bb =
        get_times(size, s2 + s2, s2.size() - size + 1, s2.size() - 1);
    matrix now = power(stand, N - numnow - 2);
    int num_a = 0;
    C[0] = 0, C[1] = 1, C[2] = 0, C[3] = 0;
    for (int j = 0; j < 4; ++j)
      num_a = (num_a + 1LL * now.M[j][0] * C[j]) % MOD;
    int num_b = 0;
    C[0] = 1, C[1] = 0, C[2] = 0, C[3] = 0;
    for (int j = 0; j < 4; ++j)
      num_b = (num_b + 1LL * now.M[j][0] * C[j]) % MOD;
    int num_ab = 0;
    C[0] = 0, C[1] = 0, C[2] = 0, C[3] = 1;
    for (int j = 0; j < 4; ++j)
      num_ab = (num_ab + 1LL * now.M[j][0] * C[j]) % MOD;
    now = power(stand, N - numnow - 3);
    int num_ba = 0;
    C[0] = 1, C[1] = 0, C[2] = 0, C[3] = 0;
    for (int j = 0; j < 4; ++j)
      num_ba = (num_ba + 1LL * now.M[j][0] * C[j]) % MOD;
    int num_bb = 0;
    C[0] = 0, C[1] = 0, C[2] = 0, C[3] = 1;
    for (int j = 0; j < 4; ++j)
      num_bb = (num_bb + 1LL * now.M[j][0] * C[j]) % MOD;
    long long result = 1LL * times_ba * num_ba + 1LL * times_ab * num_ab +
                       1LL * times_bb * num_bb + 1LL * times_a * num_a +
                       1LL * times_b * num_b;
    while (result < 0) result += MOD;
    while (result >= MOD) result -= MOD;
    cout << result << '\n';
  }
}
