"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 trie[110][4], suf[110], tag[110], tn;
int q[110];
int f[1010][110][15];
int n, m;
inline int get(char c) {
  if (c == 'A') return 0;
  if (c == 'C') return 1;
  return c == 'G' ? 2 : 3;
}
int max(int x, int y) { return x > y ? x : y; }
int main() {
  scanf("%d %d", &n, &m);
  tn = 1;
  for (int i = 0; i < m; i++) {
    char s[15];
    scanf("%s", s);
    int j, k;
    for (j = k = 0; j < strlen(s); k = trie[k][get(s[j++])])
      if (!trie[k][get(s[j])]) {
        tag[tn] = 0;
        trie[k][get(s[j])] = tn++;
      }
    tag[k] = strlen(s);
  }
  int qn = 0;
  for (int i = 0; i < 4; i++)
    if (trie[0][i]) q[qn++] = trie[0][i];
  for (int p = 0; p < qn; p++) {
    tag[q[p]] = max(tag[q[p]], tag[suf[q[p]]]);
    for (int i = 0; i < 4; i++) {
      int next = trie[q[p]][i];
      if (next) {
        suf[next] = trie[suf[q[p]]][i];
        q[qn++] = next;
      } else
        trie[q[p]][i] = trie[suf[q[p]]][i];
    }
  }
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < tn; j++)
      for (int k = 0; k < 12; k++)
        if (f[i][j][k]) {
          for (int c = 0; c < 4; c++) {
            int s = trie[j][c];
            if (k < tag[s]) {
              f[i + 1][s][0] += f[i][j][k];
              f[i + 1][s][0] %= 1000000009;
            } else {
              f[i + 1][s][k + 1] += f[i][j][k];
              f[i + 1][s][k + 1] %= 1000000009;
            }
          }
        }
  int ans = 0;
  for (int i = 0; i < tn; i++) ans = (ans + f[n][i][0]) % 1000000009;
  printf("%d\n", ans);
  return 0;
}
