"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 = 1000000009;
struct tv {
  int go[4];
  int link, prev, ch, depth, ff;
  bool ok;
  tv() {
    for (int i = 0; i < 4; i++) go[i] = -1;
    link = -1;
    prev = -1;
    ch = -1;
    depth = 0;
    ok = false;
    ff = -1;
  }
};
int n, m, h = 1, id[555];
string s;
tv tree[111];
int f[1010][111][11];
void add(int v, int pos) {
  if (pos == s.size()) {
    tree[v].ok = true;
    return;
  }
  int ch = id[s[pos]];
  if (tree[v].go[ch] == -1) {
    tree[h].prev = v;
    tree[h].depth = tree[v].depth + 1;
    tree[h].ch = ch;
    tree[v].go[ch] = h++;
  }
  add(tree[v].go[ch], pos + 1);
}
int getlink(int v);
int go(int v, int c);
int getlink(int v) {
  if (tree[v].link != -1) return tree[v].link;
  if (v == 0 || tree[v].prev == 0) {
    tree[v].link = 0;
    return 0;
  }
  tree[v].link = go(getlink(tree[v].prev), tree[v].ch);
  return tree[v].link;
}
int go(int v, int c) {
  if (tree[v].go[c] != -1) return tree[v].go[c];
  if (v == 0) {
    tree[v].go[c] = 0;
    return 0;
  }
  tree[v].go[c] = go(getlink(v), c);
  return tree[v].go[c];
}
int fun(int v) {
  if (tree[v].ff != -1) return tree[v].ff;
  if (v == 0) {
    tree[v].ff = 0;
    return 0;
  }
  if (tree[v].ok == true) {
    tree[v].ff = tree[v].depth;
    return tree[v].ff;
  }
  tree[v].ff = fun(getlink(v));
  return tree[v].ff;
}
int main() {
  id['A'] = 0;
  id['C'] = 1;
  id['G'] = 2;
  id['T'] = 3;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> s;
    add(0, 0);
  }
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < h; j++)
      for (int k = 0; k <= 10; k++) {
        if (f[i][j][k] == 0) continue;
        for (int ch = 0; ch < 4; ch++) {
          int ii = i + 1;
          int jj = go(j, ch);
          int kk = k + 1;
          if (fun(jj) >= kk) kk = 0;
          if (kk == 11) continue;
          f[ii][jj][kk] = (f[ii][jj][kk] + f[i][j][k]) % mod;
        }
      }
  int answer = 0;
  for (int i = 0; i < h; i++) answer += f[n][i][0], answer %= mod;
  cout << answer << endl;
  return 0;
}
