"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;
int Dp[1010][110][11], id[30];
struct AcAuto {
  int next[1010][4], fail[1010], Cover[1010];
  int tot, root;
  int New_node() {
    for (int i = 0; i < 4; i++) next[tot][i] = -1;
    Cover[tot++] = 0;
    return tot - 1;
  }
  void Init() {
    tot = 0;
    root = New_node();
  }
  void Insert(char *T) {
    int len = (int)strlen(T);
    int now = root;
    for (int i = 0; i < len; i++) {
      int num = id[T[i] - 'A'];
      if (next[now][num] == -1) next[now][num] = New_node();
      now = next[now][num];
    }
    Cover[now] = len;
  }
  void Build_Failtree() {
    queue<int> Q;
    fail[root] = root;
    for (int i = 0; i < 4; i++) {
      if (next[root][i] == -1)
        next[root][i] = root;
      else {
        fail[next[root][i]] = root;
        Q.push(next[root][i]);
      }
    }
    while (!Q.empty()) {
      int now = Q.front();
      Q.pop();
      Cover[now] = max(Cover[now], Cover[fail[now]]);
      for (int i = 0; i < 4; i++) {
        if (next[now][i] == -1)
          next[now][i] = next[fail[now]][i];
        else {
          fail[next[now][i]] = next[fail[now]][i];
          Q.push(next[now][i]);
        }
      }
    }
  }
  void Solve() {
    memset(Dp, 0, sizeof(Dp));
    Dp[0][0][0] = 1;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < tot; j++) {
        for (int k = 0; k < 10; k++) {
          for (int t = 0; t < 4; t++) {
            int nowj = next[j][t], nowi = i + 1;
            if (Cover[nowj] >= k + 1)
              Dp[nowi][nowj][0] =
                  (Dp[nowi][nowj][0] + Dp[i][j][k]) % 1000000009;
            else
              Dp[nowi][nowj][k + 1] =
                  (Dp[nowi][nowj][k + 1] + Dp[i][j][k]) % 1000000009;
          }
        }
      }
    }
    int Ans = 0;
    for (int i = 0; i < tot; i++) Ans = (Ans + Dp[n][i][0]) % 1000000009;
    printf("%d\n", Ans);
    return;
  }
};
AcAuto A;
char S[20];
int main() {
  cin >> n >> m;
  id[0] = 0;
  id[2] = 1;
  id['G' - 'A'] = 2;
  id['T' - 'A'] = 3;
  A.Init();
  for (int i = 1; i <= m; i++) {
    scanf("%s", S);
    A.Insert(S);
  }
  A.Build_Failtree();
  A.Solve();
  return 0;
}
