"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 long long mod = 1000000009;
int n, m, len;
char ss[302];
long long dp[1003][102][21], ans;
int work(char ch) {
  if (ch == 'A')
    return 0;
  else if (ch == 'T')
    return 1;
  else if (ch == 'G')
    return 2;
  else
    return 3;
}
struct AC_machine {
  int siz, vl[302], fail[302], ch[302][5];
  void insert() {
    int rt = 0;
    for (int i = 0; i < len; i++) {
      int now = work(ss[i]);
      if (!ch[rt][now]) ch[rt][now] = ++siz;
      rt = ch[rt][now];
    }
    vl[rt] = len;
  }
  void getfail() {
    queue<int> q;
    for (int i = 0; i < 4; i++)
      if (ch[0][i]) q.push(ch[0][i]);
    while (!q.empty()) {
      int tmp = q.front();
      q.pop();
      vl[tmp] = max(vl[tmp], vl[fail[tmp]]);
      for (int i = 0; i < 4; i++) {
        int now = ch[tmp][i];
        if (!now) {
          ch[tmp][i] = ch[fail[tmp]][i];
          continue;
        }
        q.push(now);
        int y = fail[tmp];
        while (y && !ch[y][i]) y = fail[y];
        fail[now] = ch[y][i];
      }
    }
  }
} jr;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", ss);
    len = strlen(ss);
    jr.insert();
  }
  jr.getfail();
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= jr.siz; j++)
      for (int k = 0; k <= 10; k++)
        for (int c = 0; c < 4; c++) {
          int tmp = jr.ch[j][c];
          if (jr.vl[tmp] >= k + 1)
            dp[i + 1][tmp][0] = (dp[i + 1][tmp][0] + dp[i][j][k]) % mod;
          else
            dp[i + 1][tmp][k + 1] = (dp[i + 1][tmp][k + 1] + dp[i][j][k]) % mod;
        }
  for (int i = 0; i <= jr.siz; i++) ans = (ans + dp[n][i][0]) % mod;
  printf("%I64d\n", ans);
  return 0;
}
