"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 = 120;
const int mod = 1e9 + 9;
int val(char c) {
  switch (c) {
    case 'A':
      return 0;
    case 'T':
      return 1;
    case 'C':
      return 2;
    case 'G':
      return 3;
  }
}
char s[N];
int tar[N][4], w[N], fail[N], tot;
queue<int> q;
void ins() {
  int len = strlen(s + 1), u = 0;
  for (int i = 1; i <= len; i++) {
    int c = val(s[i]);
    if (!tar[u][c]) tar[u][c] = ++tot;
    u = tar[u][c];
  }
  w[u] = len;
}
void pre() {
  for (int i = 0; i < 4; i++)
    if (tar[0][i]) q.push(tar[0][i]);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    w[u] = max(w[u], w[fail[u]]);
    for (int c = 0; c < 4; c++)
      if (tar[u][c])
        fail[tar[u][c]] = tar[fail[u]][c], q.push(tar[u][c]);
      else
        tar[u][c] = tar[fail[u]][c];
  }
}
int dp[1010][N][12];
int dfs(int len, int p, int rst) {
  if (rst >= 10) return 0;
  if (!len) return !rst;
  if (dp[len][p][rst] != -1) return dp[len][p][rst];
  int res = 0;
  for (int c = 0; c < 4; c++)
    (res += dfs(len - 1, tar[p][c], w[tar[p][c]] > rst ? 0 : rst + 1)) %= mod;
  return dp[len][p][rst] = res;
}
int main() {
  memset(dp, -1, sizeof(dp));
  int ln, n;
  scanf("%d%d", &ln, &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    ins();
  }
  pre();
  printf("%d", dfs(ln, 0, 0));
  return 0;
}
