"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;
int next_state[110][4], max_len[110];
int dp[1010][15][110];
const int MOD = 1000000009;
int main() {
  memset(max_len, -1, sizeof(max_len));
  int N, M;
  cin >> N >> M;
  vector<string> pat(M);
  for (int i = 0; i < M; i++) {
    cin >> pat[i];
  }
  vector<string> pfx;
  for (int i = 0; i < M; i++) {
    for (int k = 0; k <= (int)pat[i].length(); k++) {
      pfx.push_back(pat[i].substr(0, k));
    }
  }
  sort(pfx.begin(), pfx.end());
  pfx.erase(unique(pfx.begin(), pfx.end()), pfx.end());
  int P = pfx.size();
  const string DNA = "AGCT";
  for (int i = 0; i < P; i++) {
    for (int k = 0; k < M; k++) {
      int lenA = pfx[i].length(), lenB = pat[k].length();
      if (lenA < lenB) continue;
      if (pfx[i].substr(lenA - lenB, lenB) == pat[k]) {
        max_len[i] = max(max_len[i], lenB);
      }
    }
    for (int c = 0; c < 4; c++) {
      string s = pfx[i] + DNA[c];
      int idx = -1;
      while (1) {
        idx = lower_bound(pfx.begin(), pfx.end(), s) - pfx.begin();
        if (idx < P && pfx[idx] == s) break;
        s = s.substr(1);
      }
      next_state[i][c] = idx;
    }
  }
  dp[0][0][0] = 1;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < 10; j++) {
      for (int k = 0; k < P; k++) {
        if (dp[i][j][k] == 0) continue;
        for (int c = 0; c < 4; c++) {
          int nk = next_state[k][c], ma = max_len[nk];
          if (j < ma) {
            (dp[i + 1][0][nk] += dp[i][j][k]) %= MOD;
          } else if (j + 1 < 10) {
            (dp[i + 1][j + 1][nk] += dp[i][j][k]) %= MOD;
          }
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < P; i++) {
    (ans += dp[N][0][i]) %= MOD;
  }
  cout << ans << endl;
  return 0;
}
