"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 AhoCorasick {
  static const int MAXN = 110;
  static const int CHARSET = 4;
  int size;
  int trie[MAXN][CHARSET];
  int fail[MAXN];
  int tag[MAXN];
  void init() {
    tag[0] = 0;
    size = 1;
    memset(trie[0], 255, sizeof(trie[0]));
  }
  void add(int len, const int* s, int t) {
    int p = 0;
    for (int i = 0; i < len; i++) {
      if (trie[p][*s] == -1) {
        tag[size] = 0;
        memset(trie[size], -1, sizeof(trie[size]));
        trie[p][*s] = size++;
      }
      p = trie[p][*s];
      s++;
    }
    tag[p] = t;
  }
  void build() {
    queue<int> que;
    fail[0] = 0;
    for (int i = 0; i < CHARSET; i++)
      if (trie[0][i] != -1) {
        fail[trie[0][i]] = 0;
        que.push(trie[0][i]);
      } else
        trie[0][i] = 0;
    while (!que.empty()) {
      int p = que.front();
      que.pop();
      tag[p] = max(tag[p], tag[fail[p]]);
      for (int i = 0; i < CHARSET; i++)
        if (trie[p][i] != -1) {
          fail[trie[p][i]] = trie[fail[p]][i];
          que.push(trie[p][i]);
        } else
          trie[p][i] = trie[fail[p]][i];
    }
  }
};
const char* DNA = "ACGT";
int n, m;
char s[15][15];
int f[1010][110][15];
AhoCorasick ac;
void input() {
  ac.init();
  int a[15], len;
  for (int i = 0; i < m; i++) {
    scanf("%s", s[i]);
    len = strlen(s[i]);
    for (int j = 0; j < len; j++)
      for (int k = 0; k < 4; k++)
        if (s[i][j] == DNA[k]) {
          a[j] = k;
          break;
        }
    ac.add(len, a, len);
  }
  ac.build();
}
void solve() {
  memset(f, 0, sizeof(f));
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < ac.size; j++)
      for (int k = 0; k <= 10; k++) {
        if (f[i][j][k] == 0) continue;
        for (int t = 0; t < 4; t++) {
          int j2 = ac.trie[j][t], k2;
          if (ac.tag[j2] >= k + 1)
            k2 = 0;
          else
            k2 = k + 1;
          if (k2 <= 10) {
            f[i + 1][j2][k2] += f[i][j][k];
            f[i + 1][j2][k2] %= 1000000009;
          }
        }
      }
  int ans = 0;
  for (int j = 0; j < ac.size; j++) ans = (ans + f[n][j][0]) % 1000000009;
  printf("%d\n", ans);
}
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    input();
    solve();
  }
  return 0;
}
