"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;
map<string, int> st;
const int maxn = 110;
int go[maxn][4];
int maxfin[maxn] = {};
const int mod = 1000000009;
int dp[1010][110][12];
vector<string> inp;
int main() {
  int n, m;
  cin >> n >> m;
  st[""];
  for (int i = 0; i < m; ++i) {
    string s;
    cin >> s;
    inp.push_back(s);
    for (int i = 0; i < (int)s.length(); ++i) st[s.substr(0, i + 1)];
  }
  int cnt = 0;
  for (map<string, int>::iterator it = st.begin(); it != st.end(); ++it, ++cnt)
    it->second = cnt;
  for (map<string, int>::iterator it = st.begin(); it != st.end(); ++it) {
    for (int ch = 0; ch < 4; ++ch) {
      string t = it->first + "ACGT"[ch];
      while (!st.count(t)) t.erase(0, 1);
      go[it->second][ch] = st[t];
    }
  }
  for (map<string, int>::iterator it = st.begin(); it != st.end(); ++it) {
    for (int i = 0; i < int((inp).size()); ++i) {
      if (int((it->first).size()) >= int((inp[i]).size()) &&
          it->first.substr(int((it->first).size()) - int((inp[i]).size())) ==
              inp[i])
        maxfin[it->second] = max(maxfin[it->second], int((inp[i]).size()));
    }
  }
  dp[0][st[""]][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < cnt; ++j) {
      for (int k = 0; k <= 10; ++k) {
        for (int ch = 0; ch < 4; ++ch) {
          int toj = go[j][ch];
          int tok = k + 1;
          if (maxfin[toj] >= tok) tok = 0;
          (dp[i + 1][toj][tok] += dp[i][j][k]) %= mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < cnt; ++i) (ans += dp[n][i][0]) %= mod;
  printf("%d\n", ans);
  return 0;
}
