"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 MAXN = 1010;
const int MAXP = 110;
const int MAXL = 21;
const int MAXD = 4;
const char DNA[4] = {'A', 'T', 'C', 'G'};
const int MO = 1000000000 + 9;
set<string> words;
string prefix[MAXP];
int pre_t;
int f[MAXN][MAXP][MAXL];
pair<int, int> g[MAXP][MAXL][4];
int n, m;
void init() {
  cin >> n >> m;
  string s;
  int len;
  prefix[0] = "";
  pre_t = 1;
  for (int i = 0; i < m; ++i) {
    cin >> s;
    words.insert(s);
    for (int j = 0; j < s.length(); ++j) prefix[pre_t++] = s.substr(0, j + 1);
  }
}
void g_cal() {
  sort(prefix, prefix + pre_t);
  pre_t = unique(prefix, prefix + pre_t) - prefix;
  string s;
  for (int i = 0; i < pre_t; ++i)
    for (int nsta, nremain, dna = 0; dna < 4; ++dna) {
      s = prefix[i] + DNA[dna];
      nsta = -1;
      for (int l = 0; l <= s.length(); ++l) {
        if (nsta != -1) break;
        for (int j = 0; j < pre_t; ++j) {
          if (s.substr(l) == prefix[j]) {
            nsta = j;
            break;
          }
        }
      }
      for (int j = 0; j <= prefix[i].length(); ++j) {
        nremain = -1;
        for (int l = 0; l < s.length() - j; ++l)
          if (words.count(s.substr(l))) {
            nremain = 0;
            break;
          }
        if (nremain == -1 && nsta != -1 && j + 1 <= prefix[nsta].length())
          nremain = j + 1;
        g[i][j][dna] = make_pair(nsta, nremain);
      }
    }
}
void solve() {
  g_cal();
  f[0][0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < pre_t; ++j)
      for (int l = 0; l <= prefix[j].length(); ++l)
        for (int dna = 0; dna < 4; ++dna)
          if (g[j][l][dna].first != -1) {
            int jj = g[j][l][dna].first, ll = g[j][l][dna].second;
            f[i][jj][ll] = (f[i - 1][j][l] + f[i][jj][ll]) % MO;
          }
  }
  int ans = 0;
  for (int i = 0; i < pre_t; ++i) ans = (ans + f[n][i][0]) % MO;
  cout << ans << endl;
}
int main() {
  init();
  solve();
  return 0;
}
