"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>
int n, m, t[105][4], p, f[105], d[105], best[105], q[105], l, h, cor[128],
    dp[1005][105][11], ans;
char s[1005];
bool w[105], v[1005][105][11];
int dfs(int x, int y, int z) {
  if (x == n) return y && d[y] == z;
  if (v[x][y][z]) return dp[x][y][z];
  v[x][y][z] = 1;
  int a = 0;
  for (int i = (0); i < (4); i++) {
    int j = y, dep = z;
    while (j && !t[j][i]) {
      dep -= d[j] - d[f[j]];
      j = f[j];
    }
    if (dep < 0 || !t[j][i]) continue;
    j = t[j][i];
    if (dep + best[j] >= d[j]) dep = d[j];
    a = (a + dfs(x + 1, j, dep)) % 1000000009;
  }
  return dp[x][y][z] = a;
}
int main() {
  cor['A'] = 0;
  cor['C'] = 1;
  cor['G'] = 2;
  cor['T'] = 3;
  scanf("%d %d", &n, &m);
  for (int i = (0); i < (m); i++) {
    scanf("%s", s);
    int r = 0;
    for (int j = 0; s[j]; j++) {
      if (!t[r][cor[s[j]]]) t[r][cor[s[j]]] = ++p;
      r = t[r][cor[s[j]]];
    }
    w[r] = 1;
  }
  q[h++] = 0;
  while (l < h) {
    int y = q[l++];
    if (w[y])
      best[y] = d[y];
    else
      best[y] = best[f[y]];
    for (int i = (0); i < (4); i++)
      if (t[y][i]) {
        if (y) {
          int x = f[y];
          while (x && !t[x][i]) x = f[x];
          if (t[x][i]) x = t[x][i];
          f[t[y][i]] = x;
        }
        d[t[y][i]] = d[y] + 1;
        q[h++] = t[y][i];
      }
  }
  printf("%d\n", dfs(0, 0, 0));
}
