"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 = 150;
const int MOD = 1000000009;
struct pp {
  int end, a[4], fail;
} node[MAXN];
int tot;
int q[MAXN];
void clr() {
  int i;
  tot++;
  for (i = 0; i < 4; i++) node[tot].a[i] = 0;
  node[tot].end = node[tot].fail = 0;
}
void ins(char s[]) {
  int temp, i;
  int h = 0;
  for (i = 0; s[i]; i++) {
    temp = s[i] - 1;
    if (node[h].a[temp] == 0) {
      clr();
      node[h].a[temp] = tot;
    }
    h = node[h].a[temp];
  }
  node[h].end = i;
}
void getfail() {
  int i;
  int f = -1, r = 0;
  int temp, p, son;
  q[0] = 0;
  while (f != r) {
    temp = q[++f];
    for (i = 0; i < 4; i++) {
      p = node[temp].fail;
      if (node[temp].a[i]) {
        son = node[temp].a[i];
        if (node[p].a[i] != son)
          node[son].fail = node[p].a[i];
        else
          node[son].fail = 0;
        q[++r] = son;
      } else
        node[temp].a[i] = node[p].a[i];
      node[temp].end = max(node[temp].end, node[p].end);
    }
  }
}
void change(char s[]) {
  int i;
  for (i = 0; s[i]; i++) {
    if (s[i] == 'A')
      s[i] = 1;
    else if (s[i] == 'C')
      s[i] = 2;
    else if (s[i] == 'G')
      s[i] = 3;
    else
      s[i] = 4;
  }
}
int dp[1005][MAXN][12];
int main() {
  int n, m, i, j, k, p;
  char str[15];
  scanf("%d%d", &n, &m);
  tot = -1;
  clr();
  for (i = 0; i < m; i++) {
    scanf("%s", str);
    change(str);
    ins(str);
  }
  getfail();
  dp[0][0][0] = 1;
  for (i = 0; i < n; i++)
    for (j = 0; j <= tot; j++)
      for (k = 0; k < 10; k++)
        for (p = 0; p < 4; p++) {
          int x = node[j].a[p];
          if (node[x].end > k) {
            dp[i + 1][x][0] += dp[i][j][k];
            dp[i + 1][x][0] %= MOD;
          } else {
            dp[i + 1][x][k + 1] += dp[i][j][k];
            dp[i + 1][x][k + 1] %= MOD;
          }
        }
  int ans = 0;
  for (i = 0; i <= tot; i++) ans = (ans + dp[n][i][0]) % MOD;
  printf("%d\n", ans);
  return 0;
}
