"Multidimensional spaces are completely out of style these days, unlike genetics problems" — thought physicist Woll and changed his subject of study to bioinformatics. Analysing results of sequencing he faced the following problem concerning DNA sequences. We will further think of a DNA sequence as an arbitrary string of uppercase letters "A", "C", "G" and "T" (of course, this is a simplified interpretation).

Let w be a long DNA sequence and s1, s2, ..., sm — collection of short DNA sequences. Let us say that the collection filters w iff w can be covered with the sequences from the collection. Certainly, substrings corresponding to the different positions of the string may intersect or even cover each other. More formally: denote by |w| the length of w, let symbols of w be numbered from 1 to |w|. Then for each position i in w there exist pair of indices l, r (1 ≤ l ≤ i ≤ r ≤ |w|) such that the substring w[l ... r] equals one of the elements s1, s2, ..., sm of the collection.

Woll wants to calculate the number of DNA sequences of a given length filtered by a given collection, but he doesn't know how to deal with it. Help him! Your task is to find the number of different DNA sequences of length n filtered by the collection {si}.

Answer may appear very large, so output it modulo 1000000009.

Input

First line contains two integer numbers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10) — the length of the string and the number of sequences in the collection correspondently. 

Next m lines contain the collection sequences si, one per line. Each si is a nonempty string of length not greater than 10. All the strings consist of uppercase letters "A", "C", "G", "T". The collection may contain identical strings.

Output

Output should contain a single integer — the number of strings filtered by the collection modulo 1000000009 (109 + 9).

Examples

Input

2 1
A


Output

1


Input

6 2
CAT
TACT


Output

2

Note

In the first sample, a string has to be filtered by "A". Clearly, there is only one such string: "AA".

In the second sample, there exist exactly two different strings satisfying the condition (see the pictures below).

<image> <image>
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000009;
const int MAXN = 128;
const int MAXK = 4;
struct TrieNode {
  int next[MAXK], fail;
  int danger;
  void init() {
    fail = -1;
    danger = 0;
    for (int i = 0; i < MAXK; ++i) next[i] = -1;
  }
} T[MAXN];
int root, Top;
void Init() {
  root = 0, Top = 1;
  T[root].init();
}
int GetId(char ch) {
  switch (ch) {
    case 'A':
      return 0;
    case 'G':
      return 1;
    case 'C':
      return 2;
    case 'T':
      return 3;
  }
}
void TrieIns(char *s) {
  int i = 0;
  int p = root;
  while (s[i]) {
    int idx = GetId(s[i]);
    if (T[p].next[idx] == -1) {
      T[Top].init();
      T[p].next[idx] = Top++;
    }
    p = T[p].next[idx];
    i++;
  }
  T[p].danger = strlen(s);
}
void Build_AC_Automation() {
  int cur, tmp;
  queue<int> q;
  T[root].fail = root;
  T[root].danger = 0;
  q.push(root);
  while (!q.empty()) {
    cur = q.front();
    q.pop();
    for (int i = 0; i < MAXK; ++i) {
      if (T[cur].next[i] != -1) {
        tmp = T[cur].next[i];
        if (cur == root)
          T[tmp].fail = root;
        else {
          T[tmp].fail = T[T[cur].fail].next[i];
          T[tmp].danger = max(T[tmp].danger, T[T[tmp].fail].danger);
        }
        q.push(tmp);
      } else {
        if (cur == root)
          T[cur].next[i] = root;
        else
          T[cur].next[i] = T[T[cur].fail].next[i];
      }
    }
  }
}
long long dp[1010][128][11];
int n, m;
int main() {
  memset(dp, 0, sizeof(dp));
  cin >> n >> m;
  Init();
  for (int i = 0; i < m; ++i) {
    char str[20];
    cin >> str;
    TrieIns(str);
  }
  Build_AC_Automation();
  dp[0][0][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < Top; ++j) {
      for (int k = 0; k <= 10; ++k) {
        if (dp[i][j][k] == 0) continue;
        for (int t = 0; t < MAXK; ++t) {
          int aa = T[j].next[t];
          if (T[aa].danger > k) {
            dp[i + 1][aa][0] += dp[i][j][k];
            dp[i + 1][aa][0] %= MOD;
          } else if (k + 1 <= 10) {
            dp[i + 1][aa][k + 1] += dp[i][j][k];
            dp[i + 1][aa][k + 1] %= MOD;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < Top; ++i) {
    ans = ans + dp[n][i][0];
    ans %= MOD;
  }
  cout << ans % MOD << endl;
}
