"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 = (int)1e9 + 9;
vector<string> q;
map<string, int> mp;
map<string, int> mp2;
char b[4] = {'A', 'T', 'G', 'C'};
int Next[105][5];
int longest[105];
int dp[105][12], dp2[105][12];
int add(int x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
  return x;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    string s;
    cin >> s;
    mp2[s] = 1;
    int L = s.length();
    for (int j = 0; j <= L; j++) q.push_back(s.substr(0, j));
  }
  sort(q.begin(), q.end());
  q.erase(unique(q.begin(), q.end()), q.end());
  int sz = 0;
  int L = q.size();
  for (int i = 0; i < L; i++) mp[q[i]] = sz++;
  for (int i = 0; i < L; i++)
    for (int j = 0; j < 4; j++) {
      string t = q[i] + b[j];
      while (!mp.count(t)) t = t.substr(1);
      Next[i][j] = mp[t];
    }
  for (int i = 0; i < L; i++)
    for (int j = 0; j < q[i].length(); j++)
      if (mp2.count(q[i].substr(j))) {
        longest[i] = (int)q[i].length() - j;
        break;
      }
  dp[0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < L; j++) {
      int mx = q[j].length();
      for (int z = 0; z <= mx; z++)
        if (dp[j][z])
          for (int k = 0; k < 4; k++) {
            int to = Next[j][k];
            int nl = z + 1;
            if (longest[to] >= nl) nl = 0;
            if (nl > q[to].length()) continue;
            dp2[to][nl] = add(dp2[to][nl], dp[j][z]);
          }
    }
    memcpy(dp, dp2, sizeof(dp));
    memset(dp2, 0, sizeof(dp2));
  }
  int ans = 0;
  for (int i = 0; i < L; i++) ans = add(ans, dp[i][0]);
  printf("%d\n", ans);
  return 0;
}
