"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;
char S[20];
int ch[150][(int)4], dp[1100][150][20], f[150], sz, val[1100], len[1100],
    last[150];
inline void init() {
  sz = 1;
  memset(ch[sz], 0, sizeof(ch[sz]));
}
inline int idx(char c) {
  if (c == 'A') return 0;
  if (c == 'T') return 1;
  if (c == 'C') return 2;
  if (c == 'G') return 3;
}
inline void insert(char *s, int v) {
  int n = strlen(s);
  int u = 0;
  for (int i = 0; i < n; i++) {
    int c = idx(s[i]);
    if (!ch[u][c]) {
      memset(ch[sz], 0, sizeof(ch[sz]));
      val[sz] = 0;
      ch[u][c] = sz++;
    }
    u = ch[u][c];
  }
  val[u] = v;
  len[u] = n;
}
inline void getFail() {
  queue<int> Q;
  f[0] = 0;
  for (int i = 0; i < (int)4; i++)
    if (ch[0][i]) Q.push(ch[0][i]);
  while (!Q.empty()) {
    int r = Q.front();
    Q.pop();
    for (int c = 0; c < (int)4; c++) {
      int u = ch[r][c];
      if (!u) {
        continue;
      }
      Q.push(u);
      int v = f[r];
      while (v && !ch[v][c]) v = f[v];
      f[u] = ch[v][c];
      last[u] = val[f[u]] ? f[u] : last[f[u]];
      len[u] = max(len[u], len[f[u]]);
    }
  }
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  init();
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= m; i++) {
    scanf("%s", S);
    insert(S, i);
  }
  getFail();
  dp[0][0][0] = 1;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= sz; j++)
      for (int k = 0; k <= 11; k++)
        if (dp[i][j][k])
          for (int c = 0; c < (int)4; c++) {
            int u = j;
            while (u && !ch[u][c]) u = f[u];
            u = ch[u][c];
            if (len[u] > k)
              dp[i + 1][u][0] = (dp[i + 1][u][0] + dp[i][j][k]) % 1000000009;
            else
              dp[i + 1][u][k + 1] =
                  (dp[i + 1][u][k + 1] + dp[i][j][k]) % 1000000009;
          }
  int ans = 0;
  for (int i = 0; i <= sz; i++) ans = (ans + dp[n][i][0]) % 1000000009;
  printf("%d\n", ans);
  return 0;
}
