"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 = 1000 + 10;
const int maxnode = 100 + 10;
long long dp[maxn][maxnode][15], n, m;
char ss[maxn];
const int mod = 1e9 + 9;
struct node {
  int ch[maxnode][5];
  int f[maxnode];
  int sz;
  int val[maxnode];
  node() {
    sz = 1;
    memset(ch[0], 0, sizeof(ch[0]));
    memset(val, 0, sizeof(val));
  }
  int idx(char ch) {
    if (ch == 'A') return 0;
    if (ch == 'G') return 1;
    if (ch == 'C') return 2;
    if (ch == 'T') return 3;
  }
  void Insert(char *s, int v) {
    int len = strlen(s);
    int u = 0;
    for (int i = 0; i < len; i++) {
      int c = idx(s[i]);
      if (!ch[u][c]) {
        memset(ch[sz], 0, sizeof(ch[sz]));
        val[sz] = 0;
        ch[u][c] = sz++;
      }
      u = ch[u][c];
    }
    val[u] = len;
  }
  void getFail() {
    queue<int> q;
    f[0] = 0;
    int u = 0;
    for (int i = 0; i < 4; i++) {
      u = ch[0][i];
      if (u) {
        q.push(u);
        f[u] = 0;
      }
    }
    while (q.size()) {
      int r = q.front();
      q.pop();
      val[r] = max(val[r], val[f[r]]);
      for (int c = 0; c < 4; c++) {
        u = ch[r][c];
        if (!u) {
          ch[r][c] = ch[f[r]][c];
          continue;
        }
        q.push(u);
        int v = f[r];
        while (v && !ch[v][c]) v = f[v];
        f[u] = ch[v][c];
      }
    }
  }
};
int main() {
  cin >> n >> m;
  node trie;
  for (int i = 0; i < m; i++) {
    scanf("%s", ss);
    int len = strlen(ss);
    trie.Insert(ss, len);
  }
  trie.getFail();
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < trie.sz; j++) {
      for (int k = 0; k <= 10; k++) {
        for (int c = 0; c < 4; c++) {
          int t = trie.ch[j][c];
          if (trie.val[t] >= k + 1) {
            dp[i + 1][t][0] = (dp[i + 1][t][0] + dp[i][j][k]) % mod;
          } else {
            dp[i + 1][t][k + 1] = (dp[i + 1][t][k + 1] + dp[i][j][k]) % mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < trie.sz; i++) {
    ans = (ans + dp[n][i][0]) % mod;
  }
  cout << ans << endl;
  return 0;
}
