"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 INF = ~0U >> 2;
const int MOD = 1000000009;
int ok[110], ch[110][4], dep[110], p[110], lnk[110], whi[110], f[1010][110][10];
char s[20];
queue<int> Q;
void update(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
int fx(char c) {
  if (c == 'A')
    return 0;
  else if (c == 'C')
    return 1;
  else if (c == 'G')
    return 2;
  else
    return 3;
}
int main() {
  int n, m;
  cin >> n >> m;
  int tot = 0;
  memset(ch, -1, sizeof(ch));
  dep[0] = 0;
  for (int i = 0; i < m; i++) {
    cin >> s;
    int t = 0;
    for (int i = 0; i < strlen(s); i++) {
      int x = fx(s[i]);
      if (ch[t][x] == -1) {
        ch[t][x] = ++tot;
        dep[tot] = dep[t] + 1;
        p[tot] = t;
        whi[tot] = x;
      }
      t = ch[t][x];
    }
    ok[t] = dep[t];
  }
  Q.push(0);
  while (!Q.empty()) {
    int t = Q.front();
    Q.pop();
    if (t == 0)
      lnk[t] = -1;
    else if (p[t] == 0)
      lnk[t] = 0;
    else {
      int k = lnk[p[t]];
      for (; k != 0 && ch[k][whi[t]] == -1; k = lnk[k])
        ;
      if (ch[k][whi[t]] != -1) k = ch[k][whi[t]];
      lnk[t] = k;
    }
    for (int i = 0; i < 4; i++)
      if (ch[t][i] != -1) Q.push(ch[t][i]);
    if (t != 0) ok[t] = max(ok[t], ok[lnk[t]]);
  }
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= tot; j++)
      for (int k = 0; k < 10; k++)
        if (f[i][j][k]) {
          for (int o = 0; o < 4; o++) {
            int t = j;
            for (; t != -1 && ch[t][o] == -1; t = lnk[t])
              ;
            if (t == -1) continue;
            t = ch[t][o];
            if (ok[t] > k)
              update(f[i + 1][t][0], f[i][j][k]);
            else if (k < 9)
              update(f[i + 1][t][k + 1], f[i][j][k]);
          }
        }
  int res = 0;
  for (int i = 0; i <= tot; i++) update(res, f[n][i][0]);
  cout << res << endl;
  return 0;
}
