"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;
long long read() {
  long long x = 0, w = 1;
  char ch = getchar();
  while (!isdigit(ch)) w = ch == '-' ? -1 : 1, ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * w;
}
const int Max_n = 1005, M = 4, mod = 1e9 + 9;
int n, m;
int a[Max_n];
char s[Max_n];
int f[Max_n][Max_n][11];
namespace Trie {
int cntT, ch[Max_n][M], sur[Max_n], len[Max_n];
void add(int Len) {
  int x = 0;
  for (int i = 0; i < Len; i++) {
    if (!ch[x][a[i]]) ch[x][a[i]] = ++cntT;
    x = ch[x][a[i]];
  }
  len[x] = max(len[x], Len);
}
void build() {
  queue<int> q;
  for (int i = 0; i < M; i++)
    if (ch[0][i]) q.push(ch[0][i]);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    len[x] = max(len[x], len[sur[x]]);
    for (int i = 0; i < M; i++)
      if (ch[x][i])
        sur[ch[x][i]] = ch[sur[x]][i], q.push(ch[x][i]);
      else
        ch[x][i] = ch[sur[x]][i];
  }
}
}  // namespace Trie
using namespace Trie;
void Mod(int &x) { x %= mod; }
int DP(int L, int x, int left) {
  int &res = f[L][x][left];
  if (left > 10) return 0;
  if (res != -1) return res;
  if (!L)
    if (!left)
      return 1;
    else
      return 0;
  res = 0;
  for (int i = 0; i < M; i++) {
    int p = ch[x][i];
    if (len[p] >= left + 1)
      Mod(res += DP(L - 1, p, 0));
    else
      Mod(res += DP(L - 1, p, left + 1));
  }
  return res;
}
int main() {
  n = read(), m = read();
  while (m--) {
    scanf("%s", s);
    for (int i = 0, lim = strlen(s); i < lim; i++)
      if (s[i] == 'A')
        a[i] = 0;
      else if (s[i] == 'C')
        a[i] = 1;
      else if (s[i] == 'T')
        a[i] = 2;
      else
        a[i] = 3;
    add(strlen(s));
  }
  build();
  memset(f, -1, sizeof(f));
  cout << DP(n, 0, 0);
}
