Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool SR(int &x) { return scanf("%d", &x) == 1; }
bool SR(long long &x) { return scanf("%lld", &x) == 1; }
bool SR(double &x) { return scanf("%lf", &x) == 1; }
bool SR(char *s) { return scanf("%s", s) == 1; }
bool RI() { return true; }
template <typename I, typename... T>
bool RI(I &x, T &...tail) {
  return SR(x) && RI(tail...);
}
void SP(const int x) { printf("%d", x); }
void SP(const long long x) { printf("%lld", x); }
void SP(const double x) { printf("%.16lf", x); }
void SP(const char *s) { printf("%s", s); }
void PL() { puts(""); }
template <typename I, typename... T>
void PL(const I x, const T... tail) {
  SP(x);
  if (sizeof...(tail)) putchar(' ');
  PL(tail...);
}
const int maxn = 1e5 + 5;
struct Node {
  static Node mem[maxn], *pmem;
  int dep;
  Node *fail, *to[2];
  int who;
  Node() : dep(-1), fail(0), to{NULL, NULL}, who(-1) {}
  Node(int _d, int _w) : dep(_d), fail(0), to{NULL, NULL}, who(_w) {}
} Node::mem[maxn], *Node::pmem = Node::mem;
vector<int> odr;
struct AC {
  AC() { root = new (Node::pmem++) Node(-1, -1); }
  Node *root;
  void add(const string &s, int id) {
    Node *pt = root;
    for (int i = 0; i < int(((int)(s).size())); i++) {
      if (!pt->to[s[i] - 'a'])
        pt->to[s[i] - 'a'] = new (Node::pmem++) Node(1 + pt->dep, -1);
      pt = pt->to[s[i] - 'a'];
    }
    pt->who = id;
  }
  void build() {
    queue<Node *> q;
    q.push(root);
    root->fail = root;
    while (((int)(q).size())) {
      Node *pt = q.front();
      q.pop();
      odr.push_back(pt - Node::mem);
      for (int i = 0; i < int(2); i++)
        if (pt->to[i]) q.push(pt->to[i]);
      if (pt->dep == -1) {
        for (int i = 0; i < int(2); i++)
          if (pt->to[i]) pt->to[i]->fail = root;
        for (int i = 0; i < int(2); i++)
          if (!pt->to[i]) pt->to[i] = root;
      } else {
        for (int i = 0; i < int(2); i++)
          if (pt->to[i]) {
            Node *&cur = pt->to[i]->fail;
            for (cur = pt->fail; cur != root && !cur->to[i]; cur = cur->fail)
              ;
            if (cur->to[i]) cur = cur->to[i];
          }
        for (int i = 0; i < int(2); i++)
          if (!pt->to[i]) {
            Node *&cur = pt->to[i];
            for (cur = pt->fail; cur != root && !cur->to[i]; cur = cur->fail)
              ;
            if (cur->to[i]) cur = cur->to[i];
          }
      }
    }
    reverse((odr).begin(), (odr).end());
  }
} ac;
const int MOD = 1e9 + 7;
const int i2 = MOD - MOD / 2;
inline int add(const int &a, const int &b) {
  return a + b < MOD ? a + b : a + b - MOD;
}
inline int mul(const int &a, const int &b) { return 1LL * a * b % MOD; }
int sig;
int dp[maxn];
void find_occur(const string &s, int ans[]) {
  memset(dp, 0, sizeof(int) * sig);
  Node *pt = ac.root;
  for (char c : s) {
    pt = pt->to[c - 'a'];
    dp[pt - Node::mem]++;
  }
  for (int i : odr)
    dp[Node::mem[i].fail - Node::mem] =
        add(dp[Node::mem[i].fail - Node::mem], dp[i]);
  for (int i = 0; i < int(sig); i++)
    if (~Node::mem[i].who) ans[Node::mem[i].who] = dp[i];
}
long long n;
int m;
string s[maxn];
void read() {
  RI(n, m);
  for (int i = 0; i < int(m); i++) {
    static char in[maxn];
    RI(in);
    s[i] = in;
  }
}
string os[maxn];
int om;
void build() {
  om = m;
  for (int i = 0; i < int(m); i++) os[i] = s[i];
  sort(s, s + m);
  m = unique(s, s + m) - s;
  for (int i = 0; i < int(m); i++) ac.add(s[i], i);
  ac.build();
  sig = Node::pmem - Node::mem;
}
array<array<int, 3>, 3> operator*(const array<array<int, 3>, 3> &a,
                                  const array<array<int, 3>, 3> &b) {
  array<array<int, 3>, 3> ret;
  for (int i = 0; i < int(3); i++)
    for (int j = 0; j < int(3); j++) ret[i][j] = 0;
  for (int i = 0; i < int(3); i++)
    for (int j = 0; j < int(3); j++)
      for (int k = 0; k < int(3); k++)
        ret[i][k] = add(ret[i][k], mul(a[i][j], b[j][k]));
  return ret;
}
int ans[maxn];
string f[maxn];
void sol_small() {
  f[1] = "a";
  f[2] = "b";
  for (int i = (3); i <= int(26); i++) f[i] = f[i - 1] + f[i - 2];
  if (n <= 26) {
    find_occur(f[n], ans);
  } else {
    static int tmp[4][maxn];
    for (int i = (27); i <= int(29); i++) f[i] = f[i - 1] + f[i - 2];
    for (int i = 0; i < int(4); i++) find_occur(f[i + 26], tmp[i]);
    array<array<int, 3>, 3> X, A;
    X[0][0] = 0;
    X[0][1] = 0;
    X[0][2] = 1;
    X[1][0] = 1;
    X[1][1] = 0;
    X[1][2] = 2;
    X[2][0] = 0;
    X[2][1] = 1;
    X[2][2] = 0;
    A[0][0] = 1;
    A[0][1] = 0;
    A[0][2] = 0;
    A[1][0] = 0;
    A[1][1] = 1;
    A[1][2] = 0;
    A[2][0] = 0;
    A[2][1] = 0;
    A[2][2] = 1;
    for (long long b = n - 26; b; b >>= 1, X = X * X)
      if (b & 1) A = A * X;
    for (int i = 0; i < int(m); i++) {
      int dlt = tmp[3][i];
      dlt = add(dlt, MOD - tmp[1][i]);
      dlt = add(dlt, MOD - tmp[1][i]);
      dlt = add(dlt, MOD - tmp[0][i]);
      dlt = mul(dlt, i2);
      int a[] = {tmp[0][i], tmp[1][i], tmp[2][i]};
      for (int j = 0; j < int(3); j++) a[j] = add(a[j], dlt);
      int res = 0;
      for (int j = 0; j < int(3); j++) res = add(res, mul(a[j], A[j][0]));
      res = add(res, MOD - dlt);
      ans[i] = res;
    }
  }
}
void sol() {
  sol_small();
  for (int i = 0; i < int(om); i++) {
    int id = lower_bound(s, s + m, os[i]) - s;
    PL(ans[id]);
  }
}
int main() {
  read();
  build();
  sol();
  return 0;
}