"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 {
  int fail, w, next[4];
} node[105];
int e;
int n, m, dp[1005][11][105];
int ref(char c) {
  if (c == 'A')
    return 0;
  else if (c == 'C')
    return 1;
  else if (c == 'G')
    return 2;
  else
    return 3;
}
void init(int idx) {
  for (int i = 0; i < 4; i++) node[idx].next[i] = 0;
  node[idx].w = node[idx].fail = 0;
}
void insert(char *str) {
  int len = strlen(str), pos = 0;
  for (int i = 0; str[i]; i++) {
    int tmp = ref(str[i]);
    if (!node[pos].next[tmp]) {
      init(e);
      node[pos].next[tmp] = e++;
    }
    pos = node[pos].next[tmp];
  }
  node[pos].w = len;
}
void bfs() {
  int head, tail, queue[105];
  head = tail = 0;
  node[0].fail = -1;
  queue[tail++] = 0;
  while (head ^ tail) {
    int u = queue[head++];
    for (int i = 0; i < 4; i++) {
      int tmp = node[u].fail;
      while (tmp != -1 && !node[tmp].next[i]) tmp = node[tmp].fail;
      if (tmp == -1)
        tmp = 0;
      else
        tmp = node[tmp].next[i];
      int &v = node[u].next[i];
      if (v) {
        queue[tail++] = v;
        node[v].fail = tmp;
        node[v].w = max(node[v].w, node[tmp].w);
      } else
        v = tmp;
    }
  }
}
int main() {
  char str[1000];
  int ans, nextlen;
  while (scanf("%d%d", &n, &m) == 2) {
    init(0);
    e = 1;
    while (m--) {
      scanf("%s", str);
      insert(str);
    }
    bfs();
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= 10; j++) {
        for (int k = 0; k < e; k++) {
          if (!dp[i][j][k]) continue;
          for (int l = 0; l < 4; l++) {
            int v = node[k].next[l];
            if (node[v].w >= j + 1)
              nextlen = 0;
            else
              nextlen = j + 1;
            if (nextlen > 10) continue;
            dp[i + 1][nextlen][v] =
                (dp[i + 1][nextlen][v] + dp[i][j][k]) % 1000000009;
          }
        }
      }
    }
    ans = 0;
    for (int i = 0; i < e; i++) ans = (ans + dp[n][0][i]) % 1000000009;
    printf("%d\n", ans);
  }
  return 0;
}
