"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 UNDEF = 0;
  static const int MAXN = 128;
  static const int CHARSET = 4;
  int end;
  int tag[MAXN];
  int fail[MAXN];
  int trie[MAXN][CHARSET];
  void init() {
    tag[0] = UNDEF;
    fill(trie[0], trie[0] + CHARSET, -1);
    end = 1;
  }
  void add(int m, const int* s, int t) {
    int p = 0;
    for (int i = 0; i < m; ++i) {
      if (trie[p][*s] == -1) {
        tag[end] = UNDEF;
        fill(trie[end], trie[end] + CHARSET, -1);
        trie[p][*s] = end++;
      }
      p = trie[p][*s];
      ++s;
    }
    tag[p] = t;
  }
  void build() {
    queue<int> bfs;
    fail[0] = 0;
    for (int i = 0; i < CHARSET; ++i) {
      if (trie[0][i] != -1) {
        fail[trie[0][i]] = 0;
        bfs.push(trie[0][i]);
      } else {
        trie[0][i] = 0;
      }
    }
    while (!bfs.empty()) {
      int p = bfs.front();
      tag[p] = max(tag[p], tag[fail[p]]);
      bfs.pop();
      for (int i = 0; i < CHARSET; ++i) {
        if (trie[p][i] != -1) {
          fail[trie[p][i]] = trie[fail[p]][i];
          bfs.push(trie[p][i]);
        } else {
          trie[p][i] = trie[fail[p]][i];
        }
      }
    }
  }
} ac;
const int TTL = 12;
const int MAXN = 1024;
const long long MOD = 1000000009LL;
long long dp[MAXN][AhoCorasick::MAXN][TTL];
int main() {
  int n, m, l;
  int a[80];
  char buf[80];
  long long ans;
  const char pat[] = "ATCG";
  scanf("%d%d", &n, &m);
  ac.init();
  for (int i = 0; i < m; ++i) {
    scanf("%s", buf);
    l = strlen(buf);
    for (int i = 0; i < l; ++i) {
      a[i] = strchr(pat, buf[i]) - pat;
    }
    ac.add(l, a, l);
  }
  ac.build();
  dp[0][0][0] = 1LL;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < ac.end; ++j) {
      for (int k = 0; k < TTL; ++k) {
        if (dp[i][j][k] == 0) {
          continue;
        }
        for (int t = 0; t < 4; ++t) {
          int jj = ac.trie[j][t];
          if (ac.tag[jj] > k) {
            dp[i + 1][jj][0] += dp[i][j][k];
            dp[i + 1][jj][0] %= MOD;
          } else if (k + 1 < TTL) {
            dp[i + 1][jj][k + 1] += dp[i][j][k];
            dp[i + 1][jj][k + 1] %= MOD;
          }
        }
      }
    }
  }
  ans = 0;
  for (int j = 0; j < ac.end; ++j) {
    ans += dp[n][j][0];
  }
  printf("%d\n", (int)(ans % MOD));
  return 0;
}
