"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 char pat[] = "ACGT";
inline int get(char c) { return strchr(pat, c) - pat; }
struct Aho_Corasick {
  int N;
  int maxlen[128];
  int fail[128];
  int trie[128][4];
  inline void init() {
    maxlen[0] = 0;
    fill(trie[0], trie[0] + 4, -1);
    N = 1;
  }
  inline void insert(string s) {
    int m = s.size();
    int p = 0;
    for (int i = 0; i < m; ++i) {
      int c = get(s[i]);
      if (trie[p][c] == -1) {
        trie[p][c] = N;
        fill(trie[N], trie[N] + 4, -1);
        maxlen[N] = 0;
        ++N;
      }
      p = trie[p][c];
    }
    maxlen[p] = m;
  }
  inline void build() {
    queue<int> q;
    fail[0] = 0;
    for (int i = 0; i < 4; ++i) {
      if (trie[0][i] != -1)
        fail[trie[0][i]] = 0, q.push(trie[0][i]);
      else
        trie[0][i] = 0;
    }
    while (!q.empty()) {
      int p = q.front();
      q.pop();
      maxlen[p] = max(maxlen[p], maxlen[fail[p]]);
      for (int i = 0; i < 4; ++i) {
        if (trie[p][i] != -1)
          fail[trie[p][i]] = trie[fail[p]][i], q.push(trie[p][i]);
        else
          trie[p][i] = trie[fail[p]][i];
      }
    }
  }
};
Aho_Corasick AC;
const long long MOD = 1e9 + 9;
inline void Add(long long &x, long long y) {
  x += y;
  if (x < 0)
    x += MOD;
  else if (x >= MOD)
    x -= MOD;
  return;
}
long long dp[1024][128][12];
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  string s;
  long long ans;
  cin >> n >> m;
  AC.init();
  for (int i = 0; i < m; ++i) {
    cin >> s;
    AC.insert(s);
  }
  AC.build();
  dp[0][0][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < AC.N; ++j) {
      for (int k = 0; k < 12; ++k) {
        if (!dp[i][j][k]) continue;
        for (int c = 0; c < 4; ++c) {
          int jj = AC.trie[j][c];
          if (AC.maxlen[jj] > k)
            Add(dp[i + 1][jj][0], dp[i][j][k]);
          else if (k + 1 < 12)
            Add(dp[i + 1][jj][k + 1], dp[i][j][k]);
        }
      }
    }
  }
  ans = 0;
  for (int j = 0; j < AC.N; ++j) Add(ans, dp[n][j][0]);
  cout << ans;
  return 0;
}
