"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 = 1e9 + 9;
int f[1001][1001][11], n, m, ch[1001][4], val[1001], fail[1001], mx, tot;
string ss;
void add(int &x, int y) {
  x += y;
  if (x < 0) x += mod;
  while (x >= mod) x -= mod;
}
int pos(char c) {
  if (c == 'A') return 0;
  if (c == 'C') return 1;
  if (c == 'T')
    return 2;
  else
    return 3;
}
void insert(string s, int len) {
  int now = 0;
  for (int i = 0; i < len; i++) {
    if (ch[now][pos(s[i])] == 0) ch[now][pos(s[i])] = ++tot;
    now = ch[now][pos(s[i])];
  }
  val[now] = len;
}
void build() {
  queue<int> q;
  for (int i = 0; i <= 3; i++)
    if (ch[0][i]) fail[ch[0][i]] = 0, q.push(ch[0][i]);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i <= 3; i++) {
      int &v = ch[u][i];
      if (v == 0)
        v = ch[fail[u]][i];
      else {
        fail[v] = ch[fail[u]][i];
        q.push(v);
        val[v] = max(val[v], val[fail[v]]);
      }
    }
  }
}
int main() {
  cin >> n >> m;
  int len;
  for (int i = 1; i <= m; i++) {
    cin >> ss;
    len = ss.size();
    mx = max(mx, len);
    insert(ss, len);
  }
  build();
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= tot; j++)
      for (int k = 0, v; k <= mx; k++)
        if (v = f[i][j][k]) {
          for (int p = 0; p <= 3; p++) {
            int now = ch[j][p];
            if (val[now] > k)
              add(f[i + 1][now][0], v);
            else if (k < mx)
              add(f[i + 1][now][k + 1], v);
          }
        }
  int ans = 0;
  for (int i = 0; i <= tot; i++) add(ans, f[n][i][0]);
  cout << ans << "\n";
  return 0;
}
