"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 double pi = 3.1415926535;
const double eps = 1e-6;
queue<int> Q;
int n, m, a[20], tot, fa[110], f[1010][110][13], use[110], dang[110];
int g[110][5], suf[110];
char s[20];
int mod = 1000000009;
void update(int &a, int b) { a = (a + b) % mod; }
void tunint(char *a, int *b) {
  int le = strlen(a);
  for (int i = 1; i <= le; i++)
    if (a[i - 1] == 'A')
      b[i] = 0;
    else if (a[i - 1] == 'C')
      b[i] = 1;
    else if (a[i - 1] == 'G')
      b[i] = 2;
    else if (a[i - 1] == 'T')
      b[i] = 3;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%s", s);
    tunint(s, a);
    int le = strlen(s), now = 0;
    for (int j = 1; j <= le; j++) {
      if (g[now][a[j]] == 0) {
        g[now][a[j]] = ++tot;
        fa[tot] = now;
        use[tot] = a[j];
      }
      now = g[now][a[j]];
    }
    dang[now] = le;
  }
  Q.push(0);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    if (fa[x] == 0)
      suf[x] = 0;
    else
      suf[x] = g[suf[fa[x]]][use[x]];
    dang[x] = max(dang[x], dang[suf[x]]);
    for (int i = 0; i < 4; i++)
      if (g[x][i] == 0)
        g[x][i] = g[suf[x]][i];
      else
        Q.push(g[x][i]);
  }
  f[0][0][0] = 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= tot; j++)
      for (int len = 0; len < 10; len++)
        if (f[i - 1][j][len])
          for (int x = 0; x < 4; x++)
            if (dang[g[j][x]] >= len + 1)
              update(f[i][g[j][x]][0], f[i - 1][j][len]);
            else
              update(f[i][g[j][x]][len + 1], f[i - 1][j][len]);
  int ans = 0;
  for (int j = 0; j <= tot; j++) update(ans, f[n][j][0]);
  printf("%d\n", ans);
}
