"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>
void read(int &x) {
  x = 0;
  int flag = 1;
  char ch = ' ';
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= flag;
}
void read(long long &x) {
  x = 0;
  int flag = 1;
  char ch = ' ';
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= flag;
}
using namespace std;
char s[15];
int n, m;
struct AC_Automaton {
  int tot, mod, trie[1010][4], tag[1010], fail[1010];
  long long dp[1005][105][11];
  AC_Automaton() {
    memset(trie, -1, sizeof(trie));
    mod = 1e9 + 9;
  }
  int id(char ch) {
    if (ch == 'A') return 0;
    if (ch == 'C') return 1;
    if (ch == 'G') return 2;
    return 3;
  }
  void add(char *s, int len) {
    int u = 0;
    for (int i = 1; i <= len; i++)
      if (~trie[u][id(s[i])])
        u = trie[u][id(s[i])];
      else {
        trie[u][id(s[i])] = ++tot;
        u = tot;
      }
    tag[u] = len;
  }
  void build_AC() {
    queue<int> q;
    fail[0] = 0;
    for (int i = 0; i <= 3; i++)
      if (~trie[0][i]) {
        fail[trie[0][i]] = 0;
        q.push(trie[0][i]);
      } else
        trie[0][i] = 0;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      tag[u] = max(tag[u], tag[fail[u]]);
      for (int i = 0; i <= 3; i++)
        if (~trie[u][i]) {
          fail[trie[u][i]] = trie[fail[u]][i];
          q.push(trie[u][i]);
        } else
          trie[u][i] = trie[fail[u]][i];
    }
  }
  long long DP() {
    dp[0][0][0] = 1;
    for (int i = 0; i <= n - 1; i++)
      for (int j = 0; j <= tot; j++)
        for (int k = 0; k <= 10; k++) {
          if (dp[i][j][k] == 0) continue;
          for (int l = 0; l <= 3; l++) {
            int tmp = trie[j][l];
            if (tag[tmp] > k)
              dp[i + 1][tmp][0] = (dp[i + 1][tmp][0] + dp[i][j][k]) % mod;
            else if (k + 1 <= 10)
              dp[i + 1][tmp][k + 1] =
                  (dp[i + 1][tmp][k + 1] + dp[i][j][k]) % mod;
          }
        }
    long long ret = 0;
    for (int i = 0; i <= tot; i++) ret = (ret + dp[n][i][0]) % mod;
    return ret;
  }
} ac;
int main() {
  read(n), read(m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s + 1);
    int len = strlen(s + 1);
    ac.add(s, len);
  }
  ac.build_AC();
  printf("%lld\n", ac.DP());
  return 0;
}
