"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>
char str[15][15] = {0};
char prefix[106][15] = {0};
long o = 0;
long g[106][4] = {0}, m;
char gene[5] = "ACGT";
long f[1006][106][15] = {0};
long a[106] = {0};
char opt[15] = {0};
long find(char opt[], long l, long r) {
  long i, j, k;
  long a[15] = {0};
  for (i = 1; i <= m; i++) {
    for (j = 1; opt[j]; j++) {
      for (k = 0; opt[j + k] == str[i][1 + k] && opt[j + k] && str[i][1 + k];
           k++)
        ;
      if (!str[i][1 + k])
        for (k = 0; str[i][1 + k]; k++) a[j + k] = 1;
    }
  }
  for (j = l; j <= r; j++)
    if (!a[j]) return 0;
  return 1;
}
long search(char opt[], long l) {
  long i, j, k;
  long a[15] = {0};
  for (i = 1; i <= m; i++) {
    for (j = 1; opt[j]; j++) {
      for (k = 0; opt[j + k] == str[i][1 + k] && opt[j + k] && str[i][1 + k];
           k++)
        ;
      if (!str[i][1 + k])
        for (k = 0; str[i][1 + k]; k++) a[j + k] = 1;
    }
  }
  for (i = l + 1; a[i]; i++)
    ;
  return i - 1;
}
int main() {
  long n, i, j, k, l, t;
  long ans = 0;
  char ch;
  scanf("%ld%ld", &n, &m);
  for (i = 1; i <= m; i++) scanf("%s", str[i] + 1);
  for (i = 1; i <= m; i++)
    for (j = 1; str[i][j]; j++) {
      ch = str[i][j + 1];
      str[i][j + 1] = 0;
      for (k = 1; k <= o; k++)
        if (strcmp(prefix[k] + 1, str[i] + 1) == 0) break;
      if (k > o) {
        strcpy(prefix[++o] + 1, str[i] + 1);
        prefix[o][0] = strlen(prefix[o] + 1);
      }
      str[i][j + 1] = ch;
      if (!str[i][j + 1]) a[k] = 1;
    }
  for (i = 0; i <= o; i++) {
    for (j = 0; j < 4; j++) {
      memset(opt, 0, sizeof(opt));
      strcpy(opt, prefix[i]);
      opt[opt[0] + 1] = gene[j];
      for (k = 1; k <= prefix[i][0] + 1; k++)
        for (l = 1; l <= o; l++) {
          if (strcmp(opt + k, prefix[l] + 1) == 0) {
            g[i][j] = l;
            goto loop;
          }
        }
    loop:;
    }
  }
  f[0][0][0] = 1;
  for (i = 0; i < n; i++)
    for (j = 0; j <= o; j++)
      for (k = 0; k <= prefix[j][0]; k++)
        if (f[i][j][k]) {
          for (l = 0; l < 4; l++)
            if (g[j][l]) {
              memset(opt, 0, sizeof(opt));
              strcpy(opt, prefix[j]);
              opt[prefix[j][0] + 1] = gene[l];
              t = search(opt, k);
              if (t >= prefix[j][0] + 1 - prefix[g[j][l]][0]) {
                f[i + 1][g[j][l]]
                 [t - (prefix[j][0] + 1 - prefix[g[j][l]][0])] += f[i][j][k];
                f[i + 1][g[j][l]]
                 [t - (prefix[j][0] + 1 - prefix[g[j][l]][0])] %= 1000000009;
              }
            }
        }
  for (j = 1; j <= o; j++)
    for (k = 0; k <= prefix[j][0]; k++)
      if (find(prefix[j], k + 1, prefix[j][0])) {
        ans += f[n][j][k];
        ans %= 1000000009;
      }
  printf("%ld\n", ans);
  return 0;
}
