"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 iinf = 0x3f3f3f3f;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 1000, M = 10;
int n, m, hsh[200], et[M];
const int mod = 1e9 + 9;
void fmod(int &first) { first += first >> 31 & mod; }
const int aN = 101, C = 4;
int an, ch[aN][C], fa[aN], len[aN];
int trie() {
  for (int c(0); c < (C); ++c) ch[an][c] = -1;
  fa[an] = -1, len[an] = 0;
  return an++;
}
int rt = trie();
void insert(int i, const string &str) {
  int u = rt;
  for (const char &c : str) {
    if (!~ch[u][c]) ch[u][c] = trie();
    u = ch[u][c];
  }
  len[u] = max(len[u], int((str).size()));
}
void ac_init() {
  static queue<int> q;
  for (int c(0); c < (C); ++c)
    if (!~ch[rt][c])
      ch[rt][c] = rt;
    else
      fa[ch[rt][c]] = rt, q.push(ch[rt][c]);
  while (int((q).size())) {
    int u = q.front();
    q.pop(), len[u] = max(len[u], len[fa[u]]);
    for (int c(0); c < (C); ++c)
      if (!~ch[u][c])
        ch[u][c] = ch[fa[u]][c];
      else
        fa[ch[u][c]] = ch[fa[u]][c], q.push(ch[u][c]);
  }
}
int f[2][aN][M + 1];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  cin >> n >> m;
  hsh['A'] = 0, hsh['C'] = 1, hsh['G'] = 2, hsh['T'] = 3;
  string str;
  for (int i(0); i < (m); ++i) {
    cin >> str;
    for (char &c : str) c = hsh[c];
    insert(i, str);
  }
  ac_init();
  int a = 0;
  f[0][rt][0] = 1;
  while (n--) {
    a ^= 1;
    for (int u(0); u < (an); ++u)
      for (int j(0); j < (M + 1); ++j) f[a][u][j] = 0;
    for (int u(0); u < (an); ++u)
      for (int j(0); j < (M); ++j)
        if (f[a ^ 1][u][j])
          for (int c(0); c < (C); ++c)
            fmod(f[a][ch[u][c]][len[ch[u][c]] >= j + 1 ? 0 : j + 1] +=
                 f[a ^ 1][u][j] - mod);
  }
  int ans = 0;
  for (int u(0); u < (an); ++u) fmod(ans += f[a][u][0] - mod);
  cout << ans << '\n';
  return 0;
}
