"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 N = 1004, MaxQ = N - 1, INF = 1000000009;
int f[N][11][201], t[N][5], len[N], q[N], fail[N];
int n, m, Size;
char s[N];
int calc(char c) {
  if (c == 'A') return 0;
  if (c == 'C') return 1;
  if (c == 'G') return 2;
  if (c == 'T') return 3;
}
void prepare() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%s", s);
    int L = strlen(s);
    int r = 0;
    for (int j = 0; j < L; ++j) {
      int ch = calc(s[j]);
      if (!t[r][ch]) t[r][ch] = ++Size;
      r = t[r][ch];
    }
    len[r] = L;
  }
}
void makefail() {
  int head = 0, tail = 1;
  q[1] = 0;
  while (head != tail) {
    head = head % MaxQ + 1;
    int r = q[head];
    for (int i = 0; i < 4; ++i) {
      if (!t[r][i]) continue;
      if (r) {
        int p = fail[r];
        while (p) {
          if (t[p][i]) break;
          p = fail[p];
        }
        fail[t[r][i]] = t[p][i];
      }
      if (len[fail[t[r][i]]] > len[t[r][i]]) len[t[r][i]] = len[fail[t[r][i]]];
      tail = tail % MaxQ + 1;
      q[tail] = t[r][i];
    }
  }
}
int run(int x, int i) {
  while (x && !t[x][i]) x = fail[x];
  x = t[x][i];
  return x;
}
void solve() {
  f[0][0][0] = 1;
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= 10; ++j)
      for (int k = 0; k <= Size; ++k) {
        if (!f[i][j][k]) continue;
        for (int t = 0; t < 4; ++t) {
          int w = run(k, t);
          int j1 = j + 1;
          if (j1 <= len[w]) j1 = 0;
          if (j1 > 10) continue;
          f[i + 1][j1][w] += f[i][j][k];
          if (f[i + 1][j1][w] >= INF) f[i + 1][j1][w] -= INF;
        }
      }
  int ans = 0;
  for (int k = 0; k <= Size; ++k) ans = (ans + f[n][0][k]) % INF;
  printf("%d\n", ans);
}
int main() {
  prepare();
  makefail();
  solve();
}
