"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 long long N = 1e3 + 5, MOD = 1e9 + 9;
long long dp[N][N][15];
map<char, long long> m;
long long arr[N][4], fail[N], cnt[N], pos;
void insert(string &s) {
  long long now = 0;
  for (auto c : s) {
    long long &to = arr[now][m[c]];
    now = to ? to : to = ++pos;
  }
  cnt[now] = s.size();
}
void build() {
  queue<long long> q;
  for (long long i = 0; i < 4; i++)
    if (arr[0][i]) q.push(arr[0][i]);
  while (!q.empty()) {
    long long now = q.front();
    q.pop();
    for (long long i = 0; i < 4; i++) {
      long long &to = arr[now][i], tmp;
      if (to == 0)
        to = arr[fail[now]][i];
      else {
        q.push(to), tmp = fail[now];
        while (tmp && !arr[tmp][i])
          cnt[to] = max(cnt[to], cnt[arr[tmp][i]]), tmp = fail[tmp];
        fail[to] = arr[tmp][i];
        cnt[to] = max(cnt[to], cnt[arr[tmp][i]]);
      }
    }
  }
}
void match(string &s) {
  long long now = 0;
  for (long long i = 0; i < s.size(); i++) {
    while (now && !arr[now][s[i] - 'a']) now = fail[now];
    if (arr[now][s[i] - 'a']) now = arr[now][s[i] - 'a'];
    if (cnt[now] > 0) {
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  m['A'] = 0, m['C'] = 1, m['G'] = 2, m['T'] = 3;
  long long n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    string s;
    cin >> s;
    insert(s);
  }
  build();
  dp[0][0][0] = 1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j <= pos; j++) {
      for (long long k = 0; k <= 10; k++) {
        if (dp[i][j][k])
          for (long long c = 0; c < 4; c++) {
            long long now = arr[j][c];
            if (cnt[now] >= k + 1)
              dp[i + 1][now][0] = (dp[i + 1][now][0] + dp[i][j][k]) % MOD;
            else
              dp[i + 1][now][k + 1] =
                  (dp[i + 1][now][k + 1] + dp[i][j][k]) % MOD;
          }
      }
    }
  }
  long long ans = 0;
  for (long long j = 0; j <= pos; j++) ans = (ans + dp[n][j][0]) % MOD;
  cout << ans << "\n";
}
