"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;
inline long long read() {
  long long Hashimoto = 0;
  bool Kanna = 1;
  char I_Love = getchar();
  while (I_Love < '0' || I_Love > '9') {
    if (I_Love == '-') Kanna = 0;
    I_Love = getchar();
  }
  while (I_Love >= '0' && I_Love <= '9') {
    Hashimoto = Hashimoto * 10 + I_Love - '0';
    I_Love = getchar();
  }
  return (Kanna ? Hashimoto : -Hashimoto);
}
template <typename T1, typename T2>
inline void Umax(T1 &a, T2 b) {
  if (a < b) a = b;
}
template <typename T1, typename T2>
inline void Umin(T1 &a, T2 b) {
  if (a > b) a = b;
}
const string FA = "ACGT";
set<string> st;
map<string, int> mp;
string pm[111];
pair<int, int> g[111][11][4];
int dp[1111][111][11];
int n, m;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 0; i < (m); ++i) {
    string s;
    cin >> s;
    st.insert(s);
    for (int i = 0; i < (s.size() + 1); ++i) {
      mp[s.substr(0, i)] = 0;
    }
  }
  int cnt = 0;
  for (map<string, int>::iterator itr = mp.begin(); itr != mp.end();
       cnt++, itr++) {
    itr->second = cnt;
    pm[cnt] = itr->first;
  }
  for (int i = 0; i < (cnt); ++i)
    for (int j = 0; j < (pm[i].size() + 1); ++j)
      for (int k = 0; k < (4); ++k) {
        g[i][j][k] = make_pair(-1, -1);
        string s = pm[i] + FA[k];
        int str = -1;
        for (int l = 0; l < (s.size() + 1); ++l) {
          if (mp.find(s.substr(l)) != mp.end()) {
            str = mp[s.substr(l)];
            break;
          }
        }
        int cov = 1;
        for (int l = 0; l < (s.size() - j); ++l) {
          if (st.find(s.substr(l)) != st.end()) {
            cov = 0;
          }
        }
        if (cov) {
          cov = j + 1;
          if (cov >= 10 || cov > pm[str].size()) {
            continue;
          }
        }
        g[i][j][k] = make_pair(str, cov);
      }
  dp[0][mp[""]][0] = 1;
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (cnt); ++j)
      for (int k = 0; k < (11); ++k)
        if (dp[i][j][k])
          for (int l = 0; l < (4); ++l) {
            pair<int, int> p = g[j][k][l];
            if (p.first == -1) continue;
            dp[i + 1][p.first][p.second] =
                (dp[i + 1][p.first][p.second] + dp[i][j][k]) % 1000000009;
          }
  int ans = 0;
  for (int i = 0; i < (cnt); ++i) {
    ans = (ans + dp[n][i][0]) % 1000000009;
  }
  cout << ans << endl;
  ;
  return 0;
}
