"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;
const int mod = 1000000009;
const char *p = "ACGT";
int TO[110][4];
int cut[110][4];
map<string, int> SET;
int n, m, cnt, ans;
string s[10];
int dp[1005][110][11];
void init() {
  cnt = 0;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j <= s[i].size(); j++) {
      string t = s[i].substr(0, j);
      if (SET.find(t) == SET.end()) SET[t] = cnt++;
    }
  }
  for (map<string, int>::iterator it = SET.begin(); it != SET.end(); it++) {
    string ss = it->first;
    int id = it->second;
    for (int i = 0; i < 4; i++) {
      string t = ss + p[i];
      bool ok = false;
      for (int j = 0; j < t.length(); j++) {
        string h = t.substr(j, t.length() - j);
        if (SET.find(h) != SET.end()) {
          ok = true;
          TO[id][i] = SET[h];
          break;
        }
      }
      if (!ok) TO[id][i] = SET[""];
      cut[id][i] = 0;
      for (int j = 0; j < t.length(); j++) {
        string h = t.substr(j, t.length() - j);
        bool ok = false;
        for (int k = 0; k < m; k++)
          if (h == s[k]) {
            cut[id][i] = t.length() - j;
            ok = true;
          }
        if (ok) break;
      }
    }
  }
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) cin >> s[i];
  init();
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < cnt; j++)
      for (int reach = 0; reach <= 10; reach++) {
        if (dp[i][j][reach] == 0) continue;
        for (int k = 0; k < 4; k++) {
          int newreach = cut[j][k] >= reach + 1 ? 0 : reach + 1;
          if (newreach > 10) continue;
          dp[i + 1][TO[j][k]][newreach] += dp[i][j][reach];
          dp[i + 1][TO[j][k]][newreach] %= mod;
        }
      }
  int ans = 0;
  for (int i = 0; i < cnt; i++) {
    ans += dp[n][i][0];
    ans %= mod;
  }
  printf("%d\n", ans);
  return 0;
}
