"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, fail[12][12], tam[12];
char mat[12][12];
map<pair<pair<int, int>, vector<int> >, int> mem;
char K[] = "ACGT";
void buildFail(int k, char *p) {
  int m = tam[k] = strlen(p);
  int j = fail[k][0] = -1;
  for (int i = 1; i <= m; i++) {
    while (j >= 0 && p[j] != p[i - 1]) j = fail[k][j];
    fail[k][i] = ++j;
  }
}
int solve(int last, int ind, vector<int> v) {
  if (ind == n) return last == n - 1;
  if (mem.find(pair<pair<int, int>, vector<int> >(pair<int, int>(last, ind),
                                                  v)) != mem.end())
    return mem[pair<pair<int, int>, vector<int> >(pair<int, int>(last, ind),
                                                  v)];
  int &r =
      mem[pair<pair<int, int>, vector<int> >(pair<int, int>(last, ind), v)];
  r = 0;
  for (int i = 0; i < 4; i++) {
    char ch = K[i];
    vector<int> vv = v;
    int mx = 0;
    bool ok = false;
    for (int j = 0; j < m; j++) {
      int k = vv[j];
      if (k == tam[j]) k = fail[j][k];
      while (k >= 0 && mat[j][k] != ch) k = fail[j][k];
      vv[j] = ++k;
      mx = max(mx, k);
      if (k == tam[j] && k >= ind - last) ok = true;
    }
    if (mx < ind - last) continue;
    if (!ok)
      r += solve(last, ind + 1, vv);
    else
      r += solve(ind, ind + 1, vv);
    if (r >= mod) r -= mod;
  }
  return r;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf(" %s", mat[i]);
    buildFail(i, mat[i]);
  }
  vector<int> v(m, 0);
  printf("%d\n", solve(-1, 0, v));
  return 0;
}
