"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>
const int MOD = 1000000009;
const char genetic[4] = {'A', 'C', 'G', 'T'};
int n, m, nodeCnt, depth[111], childs[111][4], fail[111], transform[111][4],
    queue[111], dp[1111][111][11];
bool terminated[111];
int charToInt(char c) {
  for (int i = 0; i < 4; ++i) {
    if (c == genetic[i]) {
      return i;
    }
  }
  return -1;
}
void update(int &x, int a) { x = (x + a) % MOD; }
int main() {
  scanf("%d%d", &n, &m);
  nodeCnt = 1;
  memset(childs, 0, sizeof(childs));
  memset(terminated, 0, sizeof(terminated));
  for (int i = 0; i < m; ++i) {
    char str[11];
    scanf("%s", str);
    int len = strlen(str), cur = 0;
    for (int j = 0; j < len; ++j) {
      depth[cur] = j;
      int c = charToInt(str[j]);
      if (not childs[cur][c]) {
        childs[cur][c] = nodeCnt++;
      }
      cur = childs[cur][c];
    }
    depth[cur] = len;
    terminated[cur] = true;
  }
  int head = 0, tail = 0;
  for (int c = 0; c < 4; ++c) {
    transform[0][c] = childs[0][c];
    if (childs[0][c]) {
      fail[childs[0][c]] = 0;
      queue[tail++] = childs[0][c];
    }
  }
  while (head != tail) {
    int cur = queue[head++];
    if (terminated[fail[cur]] and not terminated[cur]) {
      terminated[cur] = true;
      depth[cur] = depth[fail[cur]];
    }
    for (int c = 0; c < 4; ++c) {
      if (childs[cur][c]) {
        transform[cur][c] = childs[cur][c];
        fail[childs[cur][c]] = transform[fail[cur]][c];
        queue[tail++] = childs[cur][c];
      } else {
        transform[cur][c] = transform[fail[cur]][c];
      }
    }
  }
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < nodeCnt; ++j) {
      for (int k = 0; k < 11; ++k) {
        if (dp[i][j][k]) {
          for (int c = 0; c < 4; ++c) {
            int next = transform[j][c], ex = k + 1;
            if (terminated[next]) {
              if (k < depth[next]) {
                ex = 0;
              }
            }
            if (ex >= 11) {
              continue;
            }
            update(dp[i + 1][next][ex], dp[i][j][k]);
          }
        }
      }
    }
  }
  int result = 0;
  for (int j = 0; j < nodeCnt; ++j) {
    update(result, dp[n][j][0]);
  }
  printf("%d\n", result);
  return 0;
}
