"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;
long long n, m, c, suf[111], nx[111][4], dp[1010][111][11];
string s[11], C = "ATCG";
set<string> S;
map<string, int> M;
map<string, int>::iterator it;
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> s[i];
    S.insert(s[i]);
    for (int j = 0; j < s[i].size() + 1; j++)
      if (M.find(s[i].substr(0, j)) == M.end()) {
        M[s[i].substr(0, j)] = c;
        c++;
      }
  }
  for (it = M.begin(); it != M.end(); it++) {
    for (int i = 0; i < it->first.size() + 1; i++)
      if (S.count(it->first.substr(i))) {
        suf[it->second] = it->first.size() - i;
        break;
      }
  }
  for (it = M.begin(); it != M.end(); it++)
    for (int i = 0; i < 4; i++) {
      string t = it->first + C[i];
      while (M.find(t) == M.end()) t = t.substr(1);
      nx[it->second][i] = M[t];
    }
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < c; j++)
      for (int k = 0; k < 10; k++)
        if (dp[i][j][k])
          for (int l = 0; l < 4; l++) {
            int next = nx[j][l];
            if (suf[next] >= k + 1)
              dp[i + 1][next][0] =
                  (dp[i + 1][next][0] + dp[i][j][k]) % 1000000009LL;
            else {
              if (k + 1 == 10) continue;
              dp[i + 1][next][k + 1] =
                  (dp[i + 1][next][k + 1] + dp[i][j][k]) % 1000000009LL;
            }
          }
  long long ans = 0;
  for (int i = 0; i < c; i++) ans = (ans + dp[n][i][0]) % 1000000009LL;
  cout << ans;
}
