"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;
struct trie {
  int mx, id;
  trie *fail, *next[4];
  trie() {
    fail = NULL;
    for (int i = 0; i < 4; i++) next[i] = NULL;
    mx = id = 0;
  }
} e[10010], *q[10010];
int cnt = 0;
trie *root = &e[cnt++];
char s[15];
int change(char c) {
  if (c == 'A') return 0;
  if (c == 'C') return 1;
  if (c == 'G') return 2;
  return 3;
}
void insert() {
  trie *p = root;
  int i = 0, idx;
  while (s[i]) {
    idx = change(s[i++]);
    if (!p->next[idx]) {
      e[cnt].id = cnt;
      p->next[idx] = &e[cnt++];
    }
    p = p->next[idx];
  }
  p->mx = i;
}
void build_ac_automation() {
  root->fail = root;
  q[0] = root;
  int top = 0, tail = 1;
  while (top < tail) {
    trie *p = q[top++];
    p->mx = max(p->mx, p->fail->mx);
    for (int i = 0; i < 4; i++) {
      if (!p->next[i]) {
        if (p == root)
          p->next[i] = root;
        else
          p->next[i] = p->fail->next[i];
      } else {
        if (p == root)
          p->next[i]->fail = root;
        else
          p->next[i]->fail = p->fail->next[i];
        q[tail++] = p->next[i];
      }
    }
  }
}
const int mod = 1000000009;
int dp[1010][110][12], n, m;
void solve() {
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < cnt; j++) {
      for (int k = 0; k < 10; k++) {
        for (int r = 0; r < 4; r++) {
          if (!e[j].next[r]) continue;
          int tmp = e[j].next[r]->id;
          if (e[tmp].mx > k) {
            dp[i + 1][tmp][0] = (dp[i + 1][tmp][0] + dp[i][j][k]) % mod;
          } else
            dp[i + 1][tmp][k + 1] = (dp[i + 1][tmp][k + 1] + dp[i][j][k]) % mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < cnt; i++) ans = (ans + dp[n][i][0]) % mod;
  printf("%d\n", ans);
}
int main() {
  scanf("%d%d", &n, &m);
  while (m--) {
    scanf("%s", s);
    insert();
  }
  build_ac_automation();
  solve();
  return 0;
}
