"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 M = 1000000009;
int n, m, p, q, node = 0, l, r, len, ans = 0, u;
char s[105], a[105], Q[105];
int ch[105][5], en[105], fail[105], level[105];
int f[1005][105][15];
int w[105][5];
void Add() {
  p = 0;
  for (int i = 1; i <= len; ++i) {
    if (!ch[p][a[i]]) {
      ch[p][a[i]] = ++node;
      level[node] = level[p] + 1;
    }
    p = ch[p][a[i]];
  }
  en[p] = len;
}
void Build() {
  l = 1, r = 0;
  for (int i = 1; i < 5; ++i)
    if (ch[0][i]) {
      Q[++r] = ch[0][i];
      fail[Q[r]] = 0;
    }
  for (; l <= r; ++l) {
    u = Q[l];
    for (int i = 1; i < 5; ++i)
      if (ch[u][i]) {
        p = Q[++r] = ch[u][i];
        for (q = fail[u]; q && !ch[q][i]; q = fail[q])
          ;
        if (ch[q][i])
          fail[p] = ch[q][i];
        else
          fail[p] = q;
        en[p] = max(en[p], en[fail[p]]);
      }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%s", s);
    len = strlen(s);
    for (int j = 0; j < len; ++j)
      if (s[j] == 'A')
        a[j + 1] = 1;
      else if (s[j] == 'C')
        a[j + 1] = 2;
      else if (s[j] == 'G')
        a[j + 1] = 3;
      else
        a[j + 1] = 4;
    Add();
  }
  Build();
  for (int i = 0; i <= node; ++i)
    for (int j = 1; j < 5; ++j) {
      for (p = i; p && !ch[p][j]; p = fail[p])
        ;
      if (ch[p][j])
        w[i][j] = ch[p][j];
      else
        w[i][j] = p;
    }
  f[0][0][0] = 1;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j <= node; ++j)
      for (int k = 0; k < 11; ++k)
        if (f[i][j][k])
          for (int t = 1; t < 5; ++t) {
            p = w[j][t];
            if (level[p] >= k + 1) {
              if (en[p] >= k + 1) {
                f[i + 1][p][0] += f[i][j][k];
                if (f[i + 1][p][0] >= M) f[i + 1][p][0] -= M;
              } else {
                f[i + 1][p][k + 1] += f[i][j][k];
                if (f[i + 1][p][k + 1] >= M) f[i + 1][p][k + 1] -= M;
              }
            }
          }
  ans = 0;
  for (int i = 0; i <= node; ++i) {
    ans += f[n][i][0];
    if (ans >= M) ans -= M;
  }
  printf("%d\n", ans);
  return 0;
}
