"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;
vector<string> P;
int n, m;
int dp[1000 + 1][120][10 + 10];
int nxt[120][4];
int match_length[120][4];
char ACGT[] = "ACGT";
int solve() {
  vector<string> pfx;
  for (int i = 0; i < P.size(); i++) {
    for (int j = 0; j <= P[i].length(); j++) {
      pfx.push_back(P[i].substr(0, j));
    }
  }
  sort(pfx.begin(), pfx.end());
  sort(P.begin(), P.end());
  pfx.erase(unique(pfx.begin(), pfx.end()), pfx.end());
  int K = pfx.size();
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < 4; j++) {
      nxt[i][j] = K;
      match_length[i][j] = 0;
      string s = pfx[i] + ACGT[j];
      int k;
      while (s.length() > 0) {
        k = lower_bound(pfx.begin(), pfx.end(), s) - pfx.begin();
        if (k < K && pfx[k] == s) {
          nxt[i][j] = k;
          break;
        }
        s = s.substr(1);
      }
      s = pfx[i] + ACGT[j];
      while (s.length() > 0) {
        k = lower_bound(P.begin(), P.end(), s) - P.begin();
        if (k < P.size() && P[k] == s) {
          match_length[i][j] = s.length();
          break;
        }
        s = s.substr(1);
      }
    }
  }
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int t = 0; t < n; t++) {
    for (int i = 0; i < K; i++) {
      for (int j = pfx[i].length(); j >= 0; j--) {
        if (dp[t][i][j] > 0) {
          for (int k = 0; k < 4; k++) {
            int nxt_state = nxt[i][k];
            if (nxt_state < K) {
              if (match_length[i][k] > j) {
                dp[t + 1][nxt_state][0] =
                    (dp[t + 1][nxt_state][0] + dp[t][i][j]) % 1000000009;
              } else {
                dp[t + 1][nxt_state][j + 1] =
                    (dp[t + 1][nxt_state][j + 1] + dp[t][i][j]) % 1000000009;
              }
            }
          }
        }
      }
    }
  }
  int res = 0;
  for (int i = 0; i < K; i++) {
    res = (res + dp[n][i][0]) % 1000000009;
  }
  return res;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    string s;
    cin >> s;
    P.push_back(s);
  }
  printf("%d\n", solve());
  return 0;
}
