"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>
#pragma comment(linker, "/STACK:64000000")
using namespace std;
const int mod = 1000000009;
int n, m;
string s[105];
struct vertex {
  int next[4];
  int go[4];
  int c;
  int link;
  int parent;
  int mx;
  int len;
  vertex() {
    memset(next, 255, sizeof(next));
    memset(go, 255, sizeof(go));
    link = -1;
    parent = 0;
    len = 0;
    mx = -1;
  }
};
vertex t[105];
int kv = 1;
int add(int v, int c, int len) {
  if (t[v].next[c] == -1) {
    t[kv].parent = v;
    t[kv].c = c;
    t[v].next[c] = kv;
    kv++;
  }
  v = t[v].next[c];
  t[v].len = max(t[v].len, len);
  return v;
}
int get_go(int v, int c);
int get_link(int v) {
  if (t[v].link != -1) return t[v].link;
  if (t[v].parent == 0) return t[v].link = 0;
  return t[v].link = get_go(get_link(t[v].parent), t[v].c);
}
int get_go(int v, int c) {
  if (t[v].go[c] != -1) return t[v].go[c];
  if (t[v].next[c] != -1) return t[v].go[c] = t[v].next[c];
  if (v == 0) return t[v].go[c] = 0;
  return t[v].go[c] = get_go(get_link(v), c);
}
int get_max_len(int v) {
  if (t[v].mx != -1) return t[v].mx;
  if (v == 0) return 0;
  return t[v].mx = max(t[v].len, get_max_len(get_link(v)));
}
int d[1005][105][12];
int get_num(char c) {
  if (c == 'A') return 0;
  if (c == 'G') return 1;
  if (c == 'C') return 2;
  if (c == 'T') return 3;
  assert(false);
  return -1;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < int(m); i++) {
    cin >> s[i];
    int v = 0;
    for (int j = 0; j < int(s[i].size()); j++) {
      if (j < (int)s[i].size() - 1)
        v = add(v, get_num(s[i][j]), 0);
      else
        v = add(v, get_num(s[i][j]), (int)s[i].size());
    }
  }
  d[0][0][0] = 1;
  for (int i = 0; i < int(n); i++) {
    for (int j = 0; j < int(kv); j++) {
      for (int k = 0; k < int(12); k++) {
        if (d[i][j][k] == 0) continue;
        for (int c = 0; c < int(4); c++) {
          int nj = get_go(j, c);
          int nk = k + 1;
          if (get_max_len(nj) >= nk) nk = 0;
          if (nk >= 12) continue;
          d[i + 1][nj][nk] = (d[i + 1][nj][nk] + d[i][j][k]) % mod;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < int(kv); i++) {
    ans = (ans + d[n][i][0]) % mod;
  }
  cout << ans << endl;
  return 0;
}
