"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 MOD = 1000000009;
int n, m, tail = 1;
int nxt[110][4], fail[110], ed[110], d[110], q[110];
int f[1100][110][15];
char st[15];
int ckmax(int &a, int b) { return a < b ? (a = b), 1 : 0; }
void insert() {
  int i, j;
  for (i = 0, j = 0; st[i]; i++) {
    int c = (st[i] == 'A' ? 0 : st[i] == 'C' ? 1 : st[i] == 'G' ? 2 : 3);
    if (!nxt[j][c]) d[nxt[j][c] = tail++] = d[j] + 1;
    j = nxt[j][c];
  }
  ed[j] = d[j];
}
void build() {
  int l, r, p;
  for (fail[q[l = r = 0] = 0] = 0; l <= r; l++) {
    p = q[l];
    for (int i = 0; i < 4; i++)
      !nxt[p][i]
          ? nxt[p][i] = nxt[fail[p]][i]
          : ckmax(ed[nxt[p][i]],
                  ed[fail[q[++r] = nxt[p][i]] = (p ? nxt[fail[p]][i] : 0)]);
  }
}
int F(int i, int j, int k) {
  if (~f[i][j][k]) return f[i][j][k];
  if (!i) return f[i][j][k] = ed[j] >= k;
  f[i][j][k] = 0;
  for (int a = 0; a < 4; a++)
    if (nxt[j][a]) {
      int kk = ed[j] >= k ? 1 : k + 1;
      if (d[nxt[j][a]] < kk) continue;
      (f[i][j][k] += F(i - 1, nxt[j][a], kk)) %= MOD;
    }
  return f[i][j][k];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) scanf("%s", st), insert();
  build();
  memset(f, 0xff, sizeof(f));
  printf("%d\n", F(n, 0, 0));
  return 0;
}
