"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 N = 105;
const int MOD = 1000000009;
struct Trie {
  Trie *fail, *next[26];
  int end = 0;
  void init() {
    memset(next, NULL, sizeof(next));
    fail = NULL;
    end = 0;
  }
} * q[N], *root, node[N];
int dp[1005][N][15], n, m;
int tot;
char str[N];
inline void update(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void insert(char *str) {
  Trie *cur = root;
  for (int i = 0; str[i]; i++) {
    int idx = str[i] - 'A';
    if (cur->next[idx] == NULL) {
      node[tot].init();
      cur->next[idx] = &node[tot++];
    }
    cur = cur->next[idx];
  }
  cur->end = strlen(str);
}
void init() {
  tot = 0;
  node[tot].init();
  root = &node[tot++];
}
void build() {
  int head, tail;
  head = tail = 0;
  q[0] = root;
  root->fail = NULL;
  while (head <= tail) {
    Trie *cur = q[head++];
    for (int i = 0; i < 26; i++) {
      if (cur->next[i]) {
        if (cur == root)
          cur->next[i]->fail = root;
        else {
          Trie *p = cur->fail;
          while (p) {
            if (p->next[i]) {
              cur->next[i]->fail = p->next[i];
              break;
            }
            p = p->fail;
          }
          if (p == NULL) cur->next[i]->fail = root;
          cur->next[i]->end = max(cur->next[i]->end, cur->next[i]->fail->end);
        }
        q[++tail] = cur->next[i];
      } else
        cur->next[i] = (cur == root) ? root : cur->fail->next[i];
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  init();
  for (int i = 0; i < m; i++) {
    scanf("%s", str);
    insert(str);
  }
  build();
  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++) {
        for (int a = 0; a < 26; a++) {
          int newJ = node[j].next[a] - node;
          if (node[newJ].end >= k + 1) {
            update(dp[i + 1][newJ][0], dp[i][j][k]);
          } else
            update(dp[i + 1][newJ][k + 1], dp[i][j][k]);
        }
      }
  int ans = 0;
  for (int j = 0; j < tot; j++) update(ans, dp[n][j][0]);
  printf("%d\n", ans);
  return 0;
}
