"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>
int n, m, sz, son[110][4], fail[110], term[110], par[110], l[110], q[110],
    dig[110];
long long f[1010][110][20];
long long dp(int i, int u, int o);
int main(void) {
  scanf("%d%d", &n, &m);
  memset(son, 0, sizeof(son));
  memset(term, 0, sizeof(term));
  sz = 1;
  l[1] = 0;
  int i, j, u, v, t, d;
  char str[20];
  for (i = 1; i <= m; i++) {
    scanf("%s", str);
    for (u = 1, j = 0; str[j]; j++) {
      if (str[j] == 'A')
        t = 0;
      else if (str[j] == 'C')
        t = 1;
      else if (str[j] == 'G')
        t = 2;
      else
        t = 3;
      if (!son[u][t]) {
        l[++sz] = l[u] + 1;
        son[u][t] = sz;
      }
      par[son[u][t]] = u;
      u = son[u][t];
    }
    term[u] = 1;
  }
  memset(fail, -1, sizeof(fail));
  fail[1] = 0;
  for (i = 0; i < 4; i++) fail[son[1][i]] = 1;
  int front = 1, rear = 1;
  q[1] = 1;
  while (front <= rear) {
    u = q[front];
    d = dig[front];
    front++;
    if (fail[u] < 0)
      for (fail[u] = 1, v = fail[par[u]]; v; v = fail[v])
        if (son[v][d]) {
          fail[u] = son[v][d];
          break;
        }
    for (i = 0; i < 4; i++)
      if (son[u][i]) {
        rear++;
        q[rear] = son[u][i];
        dig[rear] = i;
      }
  }
  memset(f, -1, sizeof(f));
  printf("%I64d\n", dp(0, 1, 0));
  return 0;
}
long long dp(int i, int u, int o) {
  if (f[i][u][o] >= 0)
    return f[i][u][o];
  else if (i == n)
    return f[i][u][o] = o ? 0 : 1;
  f[i][u][o] = 0;
  int j, v, w, t;
  for (j = 0; j < 4; j++) {
    for (w = 0, v = u; v && l[v] >= o; v = fail[v])
      if (son[v][j]) {
        w = son[v][j];
        break;
      }
    if (w) {
      for (t = w; l[t] >= o + 1; t = fail[t])
        if (term[t]) break;
      f[i][u][o] =
          (f[i][u][o] + dp(i + 1, w, l[t] < o + 1 ? o + 1 : 0)) % 1000000009;
    }
  }
  return f[i][u][o];
}
