"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 node {
  node *fail, *next[4];
  int flag, idx;
} trie[100];
int n, m;
int dp[1005][100][11];
int tot, front, rear;
node *q[100];
char str[15];
node *new_node() {
  node *p = &trie[tot];
  p->idx = tot;
  tot++;
  p->flag = 0;
  p->fail = NULL;
  memset(p->next, 0, sizeof(p->next));
  return p;
}
int judge(char c) {
  if (c == 'A') return 0;
  if (c == 'G') return 1;
  if (c == 'C') return 2;
  if (c == 'T') return 3;
}
void insert(char *s, node *root) {
  node *p = root;
  int len = strlen(s), idx;
  for (int i = 0; i < len; i++) {
    idx = judge(s[i]);
    if (p->next[idx] == 0) p->next[idx] = new_node();
    p = p->next[idx];
  }
  p->flag = len;
}
void setfail(node *root) {
  node *p, *temp;
  root->fail = NULL;
  q[rear++] = root;
  while (front < rear) {
    temp = q[front++];
    for (int i = 0; i < 4; i++) {
      if (temp->next[i]) {
        q[rear++] = temp->next[i];
        p = temp->fail;
        while (p && !p->next[i]) p = p->fail;
        temp->next[i]->fail = p ? p->next[i] : root;
        temp->next[i]->flag =
            max(temp->next[i]->flag, temp->next[i]->fail->flag);
      } else
        temp->next[i] = temp->fail == NULL ? root : temp->fail->next[i];
    }
  }
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    front = rear = tot = 0;
    node *root = new_node();
    while (m--) {
      scanf("%s", str);
      insert(str, root);
    }
    setfail(root);
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < tot; j++)
        for (int k = 0; k <= 10; k++) {
          if (dp[i][j][k] == 0) continue;
          for (int l = 0; l < 4; l++) {
            node *temp = &trie[j], *p;
            p = temp->next[l];
            if (p->flag > k)
              dp[i + 1][p->idx][0] =
                  (dp[i + 1][p->idx][0] + dp[i][j][k]) % 1000000009;
            else if (k < 10)
              dp[i + 1][p->idx][k + 1] =
                  (dp[i + 1][p->idx][k + 1] + dp[i][j][k]) % 1000000009;
          }
        }
    long long ret = 0;
    for (int i = 0; i < tot; i++) ret = (ret + dp[n][i][0]) % 1000000009;
    printf("%I64d\n", ret);
  }
  return 0;
}
