"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;
int ch[10010][4], tot, n, m, al[10010], ed[10010], f[1010][110][20], mp[300];
int ins(char* S) {
  int x = 0;
  for (int i = 0; S[i]; i++)
    x = (ch[x][mp[S[i]]] ? ch[x][mp[S[i]]] : ch[x][mp[S[i]]] = ++tot);
  return ed[x] = max(ed[x], (int)strlen(S)), x;
}
void byd() {
  queue<int> Q;
  for (int i = 0; i < 4; i++)
    if (ch[0][i]) Q.push(ch[0][i]);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = 0; i < 4; i++)
      if (!ch[x][i])
        ch[x][i] = ch[al[x]][i];
      else
        al[ch[x][i]] = ch[al[x]][i], Q.push(ch[x][i]),
        ed[ch[x][i]] = max(ed[ch[x][i]], ed[ch[al[x]][i]]);
  }
}
char S[10010];
signed main() {
  mp['A'] = 0, mp['C'] = 1, mp['G'] = 2, mp['T'] = 3;
  int ans = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%s", S), ins(S);
  byd();
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= tot; j++)
      for (int k = 0; k < 11; k++)
        if (f[i][j][k])
          for (int l = 0; l < 4; l++)
            f[i + 1][ch[j][l]][ed[ch[j][l]] > k ? 0 : k + 1] =
                (f[i + 1][ch[j][l]][ed[ch[j][l]] > k ? 0 : k + 1] +
                 f[i][j][k]) %
                1000000009;
  for (int i = 0; i <= tot; i++) ans = (ans + f[n][i][0]) % 1000000009;
  cout << ans;
  return 0;
}
