"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;
int ch[1005][26], n, m, tot, fail[1005], f[1005][1005][15], ed[1005];
int mp[250], mx, ans;
char s[1005];
void insert() {
  int now = 0, len = strlen(s);
  for (int i = 0; i < len; ++i) {
    if (!ch[now][mp[s[i]]]) ch[now][mp[s[i]]] = ++tot;
    now = ch[now][mp[s[i]]];
  }
  ed[now] = max(ed[now], len);
  mx = max(mx, len);
}
queue<int> q;
void build() {
  for (int i = 0; i < 4; ++i)
    if (ch[0][i]) q.push(ch[0][i]);
  while (!q.empty()) {
    int u = q.front();
    ed[u] = max(ed[fail[u]], ed[u]);
    q.pop();
    for (int i = 0; i < 4; ++i) {
      if (ch[u][i])
        q.push(ch[u][i]), fail[ch[u][i]] = ch[fail[u]][i];
      else
        ch[u][i] = ch[fail[u]][i];
    }
  }
}
inline void add(int &x, int y) { x = (long long)(x + y) % 1000000009; }
int main() {
  mp['A'] = 0, mp['T'] = 1, mp['C'] = 2, mp['G'] = 3;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) scanf("%s", s), insert();
  build();
  register int j, p, k;
  f[0][0][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (j = 0; j <= tot; ++j) {
      for (p = 0; p <= mx; ++p) {
        for (k = 0; k < 4; ++k) {
          if (ed[ch[j][k]] > p) {
            add(f[i + 1][ch[j][k]][0], f[i][j][p]);
          } else {
            add(f[i + 1][ch[j][k]][p + 1], f[i][j][p]);
          }
        }
      }
    }
  }
  for (j = 0; j <= tot; ++j) add(ans, f[n][j][0]);
  printf("%d\n", ans);
  return 0;
}
