"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 n, m;
int k;
bool debug = false;
char cc[4] = {'A', 'C', 'G', 'T'};
set<string> ori, pre;
map<string, int> rid;
int maxL, preSuff[111], mp[110][4];
long long dp[1005][11][110];
int main() {
  cin >> n >> m;
  string a;
  pre.insert("");
  for (int i = 0; i < m; i++) {
    cin >> a;
    ori.insert(a);
    int l = (int)a.size();
    for (int j = 0; j < l; j++) pre.insert(a.substr(0, j + 1));
    maxL = max(maxL, l);
  }
  int index = 0;
  for (string p : pre) {
    rid[p] = index;
    for (string o : ori) {
      auto f = p.rfind(o);
      if (f != string::npos && f == p.size() - o.size()) {
        preSuff[index] = max(preSuff[index], (int)o.size());
      }
    }
    index++;
  }
  index = 0;
  for (string p : pre) {
    for (int j = 0; j < 4; j++) {
      string x = p + cc[j];
      while (pre.find(x) == pre.end()) x = x.substr(1);
      mp[index][j] = rid[x];
    }
    index++;
  }
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < maxL; j++) {
      for (string p : pre) {
        k = rid[p];
        if (dp[i][j][k]) {
          for (int jj = 0; jj < 4; jj++) {
            string x = p + cc[jj];
            int b = preSuff[mp[k][jj]] >= j + 1 ? 0 : j + 1;
            dp[i + 1][b][mp[k][jj]] =
                (dp[i + 1][b][mp[k][jj]] + dp[i][j][k]) % 1000000009;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int k = 0; k < rid.size(); k++) ans = (ans + dp[n][0][k]) % 1000000009;
  cout << ans << endl;
  return 0;
}
