"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;
struct ACautomation {
  int chd[111][4];
  int val[111];
  int fail[111];
  int que[111];
  int dp[1005][111][12];
  int id[233];
  int cnt;
  void init() {
    fail[0] = 0;
    memset(chd[0], 0, sizeof(chd[0]));
    cnt = 1;
    id['A'] = 0, id['T'] = 1, id['C'] = 2, id['G'] = 3;
  }
  void Insert(char *a, int len) {
    int p = 0;
    for (; *a; a++) {
      if (!chd[p][id[*a]]) {
        memset(chd[cnt], 0, sizeof(chd[cnt]));
        val[cnt] = 0;
        chd[p][id[*a]] = cnt++;
      }
      p = chd[p][id[*a]];
    }
    val[p] = max(val[p], len);
  }
  void construct() {
    int *s = que, *e = que;
    for (int i = 0; i < 4; i++) {
      if (chd[0][i]) {
        fail[chd[0][i]] = 0;
        *e++ = chd[0][i];
      }
    }
    while (s != e) {
      int cur = *s++;
      val[cur] = max(val[cur], val[fail[cur]]);
      for (int i = 0; i < 4; i++) {
        int &v = chd[cur][i];
        if (v) {
          *e++ = v;
          fail[v] = chd[fail[cur]][i];
        } else
          v = chd[fail[cur]][i];
      }
    }
  }
  void solve(int n) {
    int i, j, k, z;
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    for (i = 1; i <= n; i++) {
      for (j = 0; j < cnt; j++) {
        for (k = 0; k <= 10; k++) {
          if (dp[i - 1][j][k] == 0) continue;
          for (z = 0; z < 4; z++) {
            if (val[chd[j][z]] >= k + 1)
              dp[i][chd[j][z]][0] =
                  (dp[i][chd[j][z]][0] + dp[i - 1][j][k]) % 1000000009;
            else
              dp[i][chd[j][z]][k + 1] =
                  (dp[i][chd[j][z]][k + 1] + dp[i - 1][j][k]) % 1000000009;
          }
        }
      }
    }
    int ans = 0;
    for (i = 0; i < cnt; i++) ans = (ans + dp[n][i][0]) % 1000000009;
    printf("%d\n", ans);
  }
} ac;
char str[15];
int main() {
  int n, m, len;
  scanf("%d%d", &n, &m);
  ac.init();
  while (m--) {
    scanf("%s", str);
    len = strlen(str);
    ac.Insert(str, len);
  }
  ac.construct();
  ac.solve(n);
}
