"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 = 110;
const int mod = 1e9 + 9;
struct node {
  node *next[4];
  node *fail;
  int cnt;
  void init() {
    memset(next, 0, sizeof(next));
    fail = 0;
    cnt = 0;
  }
};
node pool[maxn], *cur, *root;
int idx(char c) {
  if (c == 'A') return 0;
  if (c == 'G') return 1;
  if (c == 'T') return 2;
  if (c == 'C') return 3;
}
void insert(char *s) {
  node *p = root;
  for (int i = 0; s[i]; i++) {
    int x = idx(s[i]);
    if (p->next[x] == NULL) {
      cur->init();
      p->next[x] = cur++;
    }
    p = p->next[x];
  }
  p->cnt = strlen(s);
}
void build_fail() {
  node *p;
  queue<node *> q;
  q.push(root);
  while (!q.empty()) {
    p = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      if (p->next[i] != NULL) {
        q.push(p->next[i]);
        if (p == root)
          p->next[i]->fail = root;
        else {
          node *tmp = p->fail;
          while (tmp != NULL) {
            if (tmp->next[i] != NULL) {
              p->next[i]->fail = tmp->next[i];
              p->next[i]->cnt = max(p->next[i]->cnt, tmp->next[i]->cnt);
              break;
            }
            tmp = tmp->fail;
          }
          if (tmp == NULL) p->next[i]->fail = root;
        }
      } else {
        if (p == root)
          p->next[i] = root;
        else
          p->next[i] = p->fail->next[i];
      }
    }
  }
}
char s[maxn];
int dp[1111][maxn][15];
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    root = cur = pool;
    (cur++)->init();
    while (m--) {
      scanf("%s", s);
      insert(s);
    }
    build_fail();
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < cur - pool; j++)
        for (int k = 0; k <= 10; k++)
          if (dp[i][j][k]) {
            for (int l = 0; l < 4; l++) {
              int x = pool[j].next[l] - pool;
              if (pool[x].cnt >= k + 1)
                dp[i + 1][x][0] = (dp[i][j][k] + dp[i + 1][x][0]) % mod;
              else
                dp[i + 1][x][k + 1] = (dp[i][j][k] + dp[i + 1][x][k + 1]) % mod;
            }
          }
    int ans = 0;
    for (int i = 0; i < cur - pool; i++) ans = (ans + dp[n][i][0]) % mod;
    printf("%d\n", ans);
  }
}
