"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 maxn = 105;
const int mod = 1000000009;
int n, m, cnt;
char s[15];
int go[maxn][4], f[1005][maxn][11];
int fail[maxn], mp[maxn], end_len[maxn];
void insert(int len) {
  int now = 0;
  for (int i = 0; i < len; i++) {
    if (!go[now][mp[s[i]]]) go[now][mp[s[i]]] = ++cnt;
    now = go[now][mp[s[i]]];
  }
  end_len[now] = len;
}
void get_fail() {
  queue<int> q;
  for (int i = 0; i < 4; i++)
    if (go[0][i]) q.push(go[0][i]);
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      if (!go[now][i])
        go[now][i] = go[fail[now]][i];
      else {
        fail[go[now][i]] = go[fail[now]][i];
        end_len[go[now][i]] =
            max(end_len[go[now][i]], end_len[fail[go[now][i]]]);
        q.push(go[now][i]);
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  mp['A'] = 0, mp['G'] = 1, mp['C'] = 2, mp['T'] = 3;
  for (int i = 1; i <= m; i++) {
    scanf("%s", s);
    insert(strlen(s));
  }
  get_fail();
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= cnt; j++)
      for (int k = 0; k < 10; k++)
        for (int p = 0; p < 4; p++) {
          int now = go[j][p];
          if (end_len[now] > k)
            f[i + 1][now][0] = (f[i + 1][now][0] + f[i][j][k]) % mod;
          else
            f[i + 1][now][k + 1] = (f[i + 1][now][k + 1] + f[i][j][k]) % mod;
        }
  int ret = 0;
  for (int i = 0; i <= cnt; i++) ret = (ret + f[n][i][0]) % mod;
  printf("%d", ret);
}
