"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 n, m;
string a[15];
int go[15][15][15][4][4];
char s[5] = "ATCG";
int ret[4];
void findmatch(string &str, int prev) {
  int i, j, k;
  for (i = 0; i < m; i++) {
    if (a[i].size() <= str.size() && prev >= str.size() - a[i].size()) {
      for (k = 0, j = str.size() - a[i].size(); j < str.size(); k++, j++) {
        if (a[i][k] != str[j]) break;
      }
      if (k == a[i].size()) prev = str.size();
    }
  }
  ret[0] = ret[1] = ret[2] = ret[3] = -1;
  for (j = 0; j < str.size() && prev >= 0; j++, --prev)
    for (i = 0; i < m; i++) {
      for (k = 0; k < a[i].size() && j + k < str.size(); k++) {
        if (a[i][k] != str[j + k]) break;
      }
      if (j + k == str.size()) {
        ret[0] = i;
        ret[1] = k;
        ret[2] = prev;
        return;
      }
    }
}
void maketable(void) {
  int i, j, k, l, prev;
  memset(go, -1, sizeof(go));
  for (i = 0; i < m; i++)
    for (j = 0; j <= a[i].size(); j++) {
      for (prev = 0; prev <= j; prev++)
        for (k = 0; k < 4; k++) {
          string str = a[i].substr(0, j) + s[k];
          findmatch(str, prev);
          for (l = 0; l < 4; l++) go[i][j][prev][k][l] = ret[l];
        }
    }
}
int dp[2][15][15][15];
int main(void) {
  int i, j, k, l;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> a[i];
  }
  maketable();
  memset(dp, 0, sizeof(dp));
  int u = 0;
  const int MOD = 1000000009;
  dp[0][0][0][0] = 1;
  while (n--) {
    memset(dp[1 - u], 0, sizeof(dp[1 - u]));
    for (i = 0; i < m; i++)
      for (j = 0; j <= a[i].size(); j++)
        for (k = 0; k <= a[i].size(); k++)
          for (l = 0; l < 4; l++) {
            int *w = go[i][j][k][l];
            if (w[0] == -1) continue;
            dp[1 - u][w[0]][w[1]][w[2]] =
                (dp[1 - u][w[0]][w[1]][w[2]] + dp[u][i][j][k]) % MOD;
          }
    u = 1 - u;
  }
  int ans = 0;
  for (i = 0; i < m; i++)
    for (j = 0; j <= a[i].size(); j++) ans = (ans + dp[u][i][j][j]) % MOD;
  cout << ans << endl;
  return 0;
}
