"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 string DNA = "ACGT";
const int mod = 1000000009;
int n, m;
int good[110];
int nex[110][4];
int dp[1010][12][110];
map<string, int> mp;
int go(int pos, int rem, int past) {
  if (rem == 10) return 0;
  int &ret = dp[pos][rem][past];
  if (ret != -1) return ret;
  if (pos == n + 1) return ret = (rem == 0);
  ret = 0;
  for (int j = 0; j < 4; j++) {
    int cur = nex[past][j];
    ret += go(pos + 1, (good[cur] > rem) ? (0) : (rem + 1), cur);
    ret %= mod;
  }
  return ret;
}
int main() {
  cin >> n >> m;
  mp[""] = 0;
  for (int i = 1; i <= m; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < (int)s.size(); j++) {
      string cur = s.substr(0, j + 1);
      if (mp.find(cur) == mp.end()) {
        int sz = mp.size();
        mp[cur] = sz;
      }
    }
    good[mp[s]] = (int)s.size();
  }
  for (map<string, int>::iterator it = mp.begin(); it != mp.end(); it++) {
    for (int j = 0; j < 4; j++) {
      string tmp = (it->first) + DNA[j];
      while (mp.find(tmp) == mp.end()) tmp.erase(tmp.begin());
      nex[(it->second)][j] = mp[tmp];
    }
    string cur = (it->first);
    while (!cur.empty()) {
      cur.erase(cur.begin());
      if (mp.find(cur) != mp.end())
        good[(it->second)] = max(good[(it->second)], good[mp[cur]]);
    }
  }
  memset(dp, -1, sizeof dp);
  cout << go(1, 0, 0) << endl;
  return 0;
}
