"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>
#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
using namespace std;
class Node {
 public:
  int s[4];
  int f;
  int g;
  Node(void) : f(0), g(0) { return; }
};
char a[1020];
Node f[1020];
queue<int> q;
int g[1020 / 10][1020][11];
inline int ID(char c) { return c == 'A' ? 0 : c == 'C' ? 1 : c == 'G' ? 2 : 3; }
void InsertACAM(char *s) {
  static int c = 1;
  int i, p;
  for (i = p = 0; s[i]; i++) {
    if (!f[p].s[ID(s[i])]) f[p].s[ID(s[i])] = c++;
    p = f[p].s[ID(s[i])];
  }
  f[p].g = max(f[p].g, (signed)strlen(s));
  return;
}
void InitACAM(void) {
  int i, s, t;
  for (i = 0; i < 4; i++)
    if (f[0].s[i]) q.push(f[0].s[i]);
  for (; !q.empty(); q.pop()) {
    s = q.front();
    for (i = 0; i < 4; i++)
      if (f[s].s[i]) {
        t = f[s].s[i];
        f[t].f = f[s].f;
        while (f[t].f && !f[f[t].f].s[i]) f[t].f = f[f[t].f].f;
        f[t].f = f[f[t].f].s[i];
        f[t].g = max(f[t].g, f[f[t].f].g);
        q.push(t);
      } else
        f[s].s[i] = f[f[s].f].s[i];
  }
  return;
}
int DFS(int x, int n, int m) {
  int i;
  if (m <= f[x].g) m = 0;
  if (!n) return !m;
  if (m > 10) return 0;
  if (g[x][n][m] != -1) return g[x][n][m];
  for (i = g[x][n][m] = 0; i < 4; i++)
    g[x][n][m] = (g[x][n][m] + DFS(f[x].s[i], n - 1, m + 1)) % 1000000009;
  return g[x][n][m];
}
int main(void) {
  int n, m;
  scanf("%d %d", &n, &m);
  while (m--) {
    scanf("%s", a);
    InsertACAM(a);
  }
  InitACAM();
  memset(g, -1, sizeof(g));
  printf("%d\n", DFS(0, n, 0));
  return 0;
}
