"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>
const int MM = 500005;
const int M1 = 4;
const int mod = 1000000009;
int N, M;
char str[1005];
int tr[350];
int flag[5];
int len[25];
int L;
int q[MM], front, last;
struct node {
  int sign;
  int fail;
  int next[4];
} tree[MM];
void init(int p) {
  for (int i = 0; i < M1; i++) tree[p].next[i] = 0;
  tree[p].sign = tree[p].fail = 0;
}
void insert(char s[], int id) {
  int i = 0, p = 0, x;
  while (s[i]) {
    x = tr[s[i]];
    flag[x] = 1;
    if (tree[p].next[x] == 0) {
      tree[p].next[x] = ++L;
      init(L);
    }
    p = tree[p].next[x];
    i++;
  }
  tree[p].sign = id;
}
void build() {
  front = last = 0;
  for (int i = 0; i < M1; i++)
    if (tree[0].next[i]) {
      q[front++] = tree[0].next[i];
      tree[tree[0].next[i]].fail = 0;
    }
  while (last < front) {
    int t = q[last++];
    for (int i = 0; i < M1; ++i) {
      int tt = tree[t].next[i];
      int ff = tree[t].fail;
      if (tree[t].sign == 0) {
        tree[t].sign = tree[ff].sign;
      }
      if (tree[t].next[i]) {
        tree[tt].fail = tree[ff].next[i];
        q[front++] = tt;
      } else
        tree[t].next[i] = tree[ff].next[i];
    }
  }
}
int dp[1050][105][12];
int Gao() {
  int i, j, k, t;
  int ret = 0;
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (i = 0; i < N; i++) {
    for (j = 0; j <= L; j++) {
      for (k = 0; k <= 10; k++) {
        if (!dp[i][j][k]) continue;
        for (t = 0; t < 4; t++) {
          int tmp = tree[j].next[t];
          if (tree[tmp].sign && len[tree[tmp].sign] > k)
            dp[i + 1][tmp][0] = (dp[i + 1][tmp][0] + dp[i][j][k]) % mod;
          else
            dp[i + 1][tmp][k + 1] = (dp[i + 1][tmp][k + 1] + dp[i][j][k]) % mod;
        }
      }
    }
  }
  for (i = 0; i <= L; i++) {
    ret = (ret + dp[N][i][0]) % mod;
  }
  return ret;
}
int main() {
  tr['A'] = 0;
  tr['C'] = 1;
  tr['G'] = 2;
  tr['T'] = 3;
  scanf("%d%d", &N, &M);
  L = 0;
  init(0);
  memset(len, 0, sizeof(len));
  memset(flag, 0, sizeof(flag));
  for (int i = 1; i <= M; i++) {
    scanf("%s", str);
    len[i] = strlen(str);
    insert(str, i);
  }
  build();
  printf("%d\n", Gao());
  return 0;
}
