"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;
const int N = 12;
const int M = 1e3 + 6;
char s[N][N];
int n, trie[M][26], cnt, fail[M], len[M], m, mx,
    ind[5] = {1, 0, 'G' - 'A', 'T' - 'A', 'C' - 'A'};
const int mod = 1e9 + 9;
long long f[M][N][M];
void insert(int o) {
  int l = strlen(s[o] + 1);
  mx = max(l, mx);
  int now = 0;
  for (int i = 1, c; i <= l; i++) {
    c = s[o][i] - 'A';
    if (trie[now][c] == 0) trie[now][c] = ++cnt;
    now = trie[now][c];
  }
  len[now] = l;
}
void build() {
  queue<int> q;
  for (int i = 0; i < 26; i++)
    if (trie[0][i]) q.push(trie[0][i]);
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = 0; i < 26; i++) {
      if (trie[now][i]) {
        q.push(trie[now][i]);
        fail[trie[now][i]] = trie[fail[now]][i];
      } else
        trie[now][i] = trie[fail[now]][i];
    }
    len[now] = max(len[now], len[fail[now]]);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s[i] + 1);
    insert(i);
  }
  build();
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < mx; j++) {
      for (int k = 0; k <= cnt; k++) {
        if (!f[i][j][k]) continue;
        for (int l = 1; l <= 4; l++) {
          int now = trie[k][ind[l]];
          if (len[now] > j)
            f[i + 1][0][now] = (f[i + 1][0][now] + f[i][j][k]) % mod;
          else
            f[i + 1][j + 1][now] = (f[i + 1][j + 1][now] + f[i][j][k]) % mod;
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i <= cnt; i++) ans = (ans + f[n][0][i]) % mod;
  cout << ans;
  return 0;
}
