"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 mod = 1000000009;
struct node {
  int next[4];
  int flag;
  int f;
  int u;
} v[10005];
int ans;
char qw[4] = {'A', 'C', 'G', 'T'};
int sh[26];
char ch[1003];
int dp[1003][103][11];
void build(int x, char s[]) {
  int p = 0;
  while (s[p]) {
    int y = sh[s[p] - 'A'];
    if (v[x].next[y] != 0)
      x = v[x].next[y];
    else {
      v[x].next[y] = ans;
      x = ans++;
    }
    p++;
  }
  v[x].flag = 1;
  v[x].u = p;
}
int q[1005];
void make() {
  int top = 1;
  int tail = 0;
  q[++tail] = 0;
  while (top <= tail) {
    int x = q[top++];
    int i;
    for (i = 0; i <= 3; i++) {
      if (v[x].next[i] == 0) continue;
      int y = v[x].next[i];
      q[++tail] = y;
      int f = v[x].f;
      while (f != -1) {
        if (v[f].next[i] != 0) {
          v[y].f = v[f].next[i];
          v[y].flag |= v[v[f].next[i]].flag;
          v[y].u = max(v[y].u, v[v[f].next[i]].u);
          break;
        }
        f = v[f].f;
      }
    }
  }
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  memset(v, 0, sizeof(v));
  int i;
  ans = 1;
  for (i = 0; i <= 3; i++) {
    sh[qw[i] - 'A'] = i;
  }
  v[0].f = -1;
  for (i = 0; i <= m - 1; i++) {
    scanf("%s", ch);
    build(0, ch);
  }
  make();
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (i = 0; i <= n - 1; i++) {
    int j;
    for (j = 0; j < ans; j++) {
      int k;
      for (k = 0; k <= 9; k++) {
        int l;
        for (l = 0; l <= 3; l++) {
          int f = j;
          while (f != -1) {
            if (v[f].next[l] != 0) break;
            f = v[f].f;
          }
          if (f == -1) {
            (dp[i + 1][0][k + 1] = (dp[i + 1][0][k + 1] + dp[i][j][k]) % mod);
            continue;
          }
          int y = v[f].next[l];
          if (v[y].u > k)
            (dp[i + 1][y][0] = (dp[i + 1][y][0] + dp[i][j][k]) % mod);
          else
            (dp[i + 1][y][k + 1] = (dp[i + 1][y][k + 1] + dp[i][j][k]) % mod);
        }
      }
    }
  }
  int sum = 0;
  for (i = 0; i < ans; i++) (sum = (sum + dp[n][i][0]) % mod);
  printf("%d\n", sum);
  return 0;
}
