"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 = 1e9 + 9;
const int MAXN = 100, MAXL = 1000, MAXP = 10, ALPH = 26;
string S[MAXN + 5];
int cnt, G[MAXN + 5][ALPH + 5], fail[MAXN + 5], dep[MAXN + 5], ter[MAXN + 5];
void bfs(int v) {
  queue<int> que;
  que.push(v);
  fail[0] = -1;
  dep[0] = 0;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    ter[u] = max(ter[u], ter[fail[u]]);
    for (int i = 0; i < (int)(ALPH); i++) {
      int to = G[u][i];
      if (to == -1) continue;
      dep[to] = dep[u] + 1;
      fail[to] = u;
      do fail[to] = fail[fail[to]];
      while (fail[to] != -1 && G[fail[to]][i] == -1);
      fail[to] = fail[to] == -1 ? 0 : G[fail[to]][i];
      que.push(to);
    }
  }
}
int getnxt(int v, int ch) {
  while (v != 0 && G[v][ch] == -1) v = fail[v];
  return G[v][ch] == -1 ? 0 : G[v][ch];
}
string gen = "ACGT";
int dp[MAXL + 5][MAXN + 5][MAXP + 5];
bool occur(string S, string T) {
  for (int i = 0; i < (int)(S.length() - T.length() + 1); i++)
    if (S.substr(i, T.length()) == T) return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int l, n;
  cin >> l >> n;
  for (int i = 0; i < (int)(n); i++) cin >> S[i];
  sort(S, S + n);
  n = unique(S, S + n) - S;
  memset(G, -1, sizeof(G));
  memset(ter, -1, sizeof(ter));
  cnt = 1;
  for (int i = 0; i < (int)(n); i++) {
    int v = 0;
    for (int j = 0; j < (int)(S[i].length()); j++) {
      int ch = lower_bound(gen.begin(), gen.end(), S[i][j]) - gen.begin();
      if (G[v][ch] == -1) G[v][ch] = cnt++;
      v = G[v][ch];
    }
    ter[v] = max(ter[v], (int)S[i].length());
  }
  bfs(0);
  dp[0][0][0] = 1;
  for (int i = 0; i < (int)(l); i++)
    for (int j = 0; j < (int)(cnt); j++)
      for (int k = 0; k <= MAXP; k++)
        for (int c = 0; c < (int)(ALPH); c++) {
          int nj = getnxt(j, c);
          if (nj == 0) continue;
          int nd = ter[j] >= k ? 1 : k + 1;
          dp[i + 1][nj][nd] = (dp[i + 1][nj][nd] + dp[i][j][k]) % MOD;
        }
  int ans = 0;
  for (int i = 0; i < (int)(cnt); i++)
    if (ter[i] > 0)
      for (int j = 0; j <= ter[i]; j++) ans = (ans + dp[l][i][j]) % MOD;
  cout << ans << '\n';
  return 0;
}
