"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>
char s[10 + 1];
int ac_next[10 * 10 + 1][4], ac_fail[10 * 10 + 1], ac_no;
int ac_dist[10 * 10 + 1];
int queue[10 * 10 + 1];
int dp[2][10 * 10 + 1][10 + 1];
static inline int mapping(int c) {
  switch (c) {
    case 'A':
      return 0;
    case 'C':
      return 1;
    case 'G':
      return 2;
    case 'T':
      return 3;
    default:
      return -1;
  }
}
void ac_init(void) {
  memset(ac_next[0], 0, sizeof ac_next[0]);
  ac_dist[0] = 0;
  ac_no = 1;
}
void ac_enter(const char *s) {
  int u = 0;
  for (int i = 0; s[i] != '\0'; ++i) {
    int k = mapping(s[i]);
    if (ac_next[u][k] == 0) {
      memset(ac_next[ac_no], 0, sizeof ac_next[ac_no]);
      ac_dist[ac_no] = 0;
      ac_next[u][k] = ac_no++;
    }
    u = ac_next[u][k];
  }
  ac_dist[u] = strlen(s);
}
void ac_fix(void) {
  int s = 0, t = 0;
  ac_fail[0] = 0;
  for (int k = 0; k < 4; ++k) {
    int v = ac_next[0][k];
    if (v != 0) {
      ac_fail[v] = 0;
      queue[t++] = v;
    }
  }
  while (s < t) {
    int u = queue[s++];
    int f = ac_fail[u];
    for (int k = 0; k < 4; ++k) {
      int v = ac_next[u][k];
      if (v == 0) {
        ac_next[u][k] = ac_next[f][k];
        continue;
      }
      ac_fail[v] = ac_next[f][k];
      if (!ac_dist[v] && ac_dist[ac_fail[v]]) {
        ac_dist[v] = ac_dist[ac_fail[v]];
      }
      queue[t++] = v;
    }
  }
}
int main(void) {
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    ac_init();
    while (m--) {
      scanf("%s", s);
      ac_enter(s);
    }
    ac_fix();
    memset(dp[0], 0, sizeof(dp[0]));
    dp[0][0][0] = 1;
    for (int i = 0; i < n; ++i) {
      memset(dp[1 - (i & 1)], 0, sizeof(dp[1 - (i & 1)]));
      for (int u = 0; u < ac_no; ++u) {
        for (int j = 0; j < 10; ++j) {
          if (!dp[i & 1][u][j]) {
            continue;
          }
          for (int k = 0; k < 4; ++k) {
            int v = ac_next[u][k];
            if (j + 1 <= ac_dist[v]) {
              dp[1 - (i & 1)][v][0] =
                  (dp[1 - (i & 1)][v][0] + dp[i & 1][u][j]) % 1000000009;
            } else {
              dp[1 - (i & 1)][v][j + 1] =
                  (dp[1 - (i & 1)][v][j + 1] + dp[i & 1][u][j]) % 1000000009;
            }
          }
        }
      }
    }
    int ans = 0;
    for (int u = 0; u < ac_no; ++u) {
      if (ac_dist[u]) {
        ans = (ans + dp[n & 1][u][0]) % 1000000009;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
