"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 C = 4;
char st[20];
int ch[120][C], f[120], tmp[120][C];
int dp[1010][120][12];
int sz;
int last[120];
int getNum(int a) {
  switch (a) {
    case 'A':
      return 0;
    case 'C':
      return 1;
    case 'G':
      return 2;
    case 'T':
      return 3;
  }
}
void insert() {
  int len = strlen(st);
  int u = 0;
  for (int i = 0; i < len; i++) {
    int x = getNum(st[i]);
    if (!ch[u][x]) {
      ch[u][x] = sz++;
    }
    u = ch[u][x];
  }
  last[u] = len;
}
void getFail() {
  queue<int> q;
  f[0] = 0;
  for (int i = 0; i < C; i++)
    if (ch[0][i]) q.push(ch[0][i]);
  while (!q.empty()) {
    int r = q.front();
    q.pop();
    for (int x = 0; x < C; x++) {
      int u = ch[r][x];
      if (!u) {
        ch[r][x] = ch[f[r]][x];
        continue;
      }
      q.push(u);
      int v = f[r];
      f[u] = ch[v][x];
      last[u] = max(last[f[u]], last[u]);
    }
  }
}
int main() {
  memset(ch, 0, sizeof(ch));
  memset(last, 0, sizeof(last));
  sz = 1;
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%s", st);
    insert();
  }
  memcpy(tmp, ch, sizeof(ch));
  getFail();
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < sz; j++) {
      for (int k = 0; k < 10; k++) {
        for (int c = 0; c < C; c++) {
          int v = ch[j][c];
          if (last[v] > k)
            dp[i + 1][v][0] = (dp[i + 1][v][0] + dp[i][j][k]) % 1000000009;
          else
            dp[i + 1][v][k + 1] =
                (dp[i + 1][v][k + 1] + dp[i][j][k]) % 1000000009;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < sz; i++) {
    ans = (ans + dp[n][i][0]) % 1000000009;
  }
  printf("%d\n", ans);
}
