"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;
string dna = "ACGT";
string word[20];
set<string> words;
map<string, int> mp;
string pm[110];
pair<int, int> graph[110][20][4];
long long dp[1010][110][20];
int main() {
  ios::sync_with_stdio(false);
  int N, K, i, j, k, l, x;
  cin >> N >> K;
  for (i = 0; i < K; i++) cin >> word[i];
  for (i = 0; i < K; i++) words.insert(word[i]);
  for (i = 0; i < K; i++)
    for (j = 1; j <= word[i].length(); j++) mp[word[i].substr(0, j)] = 0;
  mp[""] = 0;
  int s = 0;
  for (auto itr = mp.begin(); itr != mp.end(); itr++) {
    (itr->second) = s;
    pm[s] = (itr->first);
    s++;
  }
  for (i = 0; i < s; i++) {
    for (j = 0; j < pm[i].length() + 1; j++) {
      for (k = 0; k < 4; k++) {
        string s = pm[i] + dna[k];
        int nextst = -1;
        for (l = 0;; l++) {
          if (mp.find(s.substr(l)) != mp.end()) {
            nextst = mp[s.substr(l)];
            break;
          }
        }
        bool cover = false;
        for (l = 0; l < pm[i].length() - j + 1; l++)
          if (words.find(s.substr(l)) != words.end()) cover = true;
        graph[i][j][k] = make_pair(-1, -1);
        int nextcover = j + 1;
        if (cover) nextcover = 0;
        if (nextcover >= 10 || nextcover > pm[nextst].length()) continue;
        graph[i][j][k] = make_pair(nextst, nextcover);
      }
    }
  }
  int start = mp[""];
  dp[0][start][0] = 1;
  for (i = 0; i < N; i++) {
    for (j = 0; j < s; j++) {
      for (k = 0; k < 10; k++) {
        if (dp[i][j][k] != 0) {
          for (x = 0; x < 4; x++) {
            pair<int, int> p = graph[j][k][x];
            if (p.first != -1) {
              int i2 = i + 1, j2 = p.first, k2 = p.second;
              dp[i2][j2][k2] = (dp[i][j][k] + dp[i2][j2][k2]) % 1000000009ll;
            }
          }
        }
      }
    }
  }
  long long ans = 0;
  for (i = 0; i < s; i++) ans = (ans + dp[N][i][0]) % 1000000009ll;
  cout << ans << endl;
  return 0;
}
