"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 maxnode = 200, sigma_size = 4, maxn = 2000, mo = 1000000000 + 9;
int sz, ch[maxnode][sigma_size], val[maxnode], f[maxnode];
int n, m, h[maxnode][maxn][20];
void init() {
  sz = 1;
  memset(ch[0], 0, sizeof(ch[0]));
  val[0] = 0;
  memset(h, -1, sizeof(h));
}
int idx(char c) {
  if (c == 'A') return 0;
  if (c == 'G') return 1;
  if (c == 'C') return 2;
  if (c == 'T') return 3;
}
void insert(char *s) {
  int l = strlen(s), u = 0;
  for (int i = 0; i < l; i++) {
    int c = idx(s[i]);
    if (!ch[u][c]) {
      ch[u][c] = sz;
      memset(ch[sz], 0, sizeof(ch[sz]));
      val[sz] = 0;
      sz++;
    }
    u = ch[u][c];
  }
  val[u] = l;
}
void get_fail() {
  queue<int> q;
  f[0] = 0;
  for (int c = 0; c < sigma_size; c++) {
    int u = ch[0][c];
    if (u) {
      q.push(u);
      f[u] = 0;
    }
  }
  while (!q.empty()) {
    int r = q.front();
    q.pop();
    for (int c = 0; c < sigma_size; c++) {
      int u = ch[r][c];
      if (!u) {
        ch[r][c] = ch[f[r]][c];
        continue;
      }
      q.push(u);
      int v = f[r];
      f[u] = ch[v][c];
      val[u] = max(val[u], val[f[u]]);
    }
  }
}
int dp(int u, int l, int k) {
  if (!l) return !k;
  if (k > 10) return 0;
  if (h[u][l][k] != -1) return h[u][l][k];
  int &ans = h[u][l][k];
  ans = 0;
  for (int c = 0; c < sigma_size; c++) {
    int v = ch[u][c];
    if (val[v] >= k + 1)
      ans += dp(v, l - 1, 0);
    else
      ans += dp(v, l - 1, k + 1);
    ans %= mo;
  }
  return ans % mo;
}
int main() {
  char s[20];
  init();
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    scanf("%s", s);
    insert(s);
  }
  get_fail();
  cout << dp(0, n, 0);
  return 0;
}
