"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 maxn = 103, mod = 1000000009;
void PlusEqual(int &x, int y) {
  if ((x += y) >= mod) x -= mod;
}
int t[maxn][4], en[maxn], fail[maxn], cnt, q[maxn], qhead, qtail, n, m,
    dp[1003][maxn][13];
char s[13];
void insert(const char *s) {
  int p = 1, i;
  for (i = 0; s[i]; i++) {
    char c = (s[i] == 'A' ? 0 : (s[i] == 'C' ? 1 : (s[i] == 'G' ? 2 : 3)));
    if (!t[p][c]) t[p][c] = ++cnt;
    p = t[p][c];
  }
  en[p] = i;
}
void build() {
  fail[1] = 0;
  for (int i = 0; i < 4; i++) t[0][i] = 1;
  qhead = 1, qtail = 0;
  q[++qtail] = 1;
  while (qhead <= qtail) {
    int p = q[qhead++];
    for (int i = 0; i < 4; i++) {
      if (t[p][i])
        fail[t[p][i]] = t[fail[p]][i],
        en[t[p][i]] = max(en[t[p][i]], en[fail[t[p][i]]]), q[++qtail] = t[p][i];
      else
        t[p][i] = t[fail[p]][i];
    }
  }
}
int query() {
  dp[0][1][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int u = 1; u <= cnt; u++) {
      for (int l = 0; l < 4; l++) {
        int v = t[u][l];
        for (int j = 0; j <= 10; j++) {
          PlusEqual(dp[i + 1][v][en[v] > j ? 0 : j + 1], dp[i][u][j]);
        }
      }
    }
  }
  int ans = 0;
  for (int u = 1; u <= cnt; u++) PlusEqual(ans, dp[n][u][0]);
  return ans;
}
int main() {
  cnt = 1;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s);
    insert(s);
  }
  build();
  printf("%d\n", query());
  return 0;
}
