"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 N = 1000 + 10;
const int M = 100 + 10;
const int P = 1000000009;
int n, m, status, i, j, k, l, num, tot, fail[M], cnt[M], dp[N][M][12],
    son[M][4], q[2000000];
bool vis[N][M][12];
string s;
map<string, int> mp;
int getId(char ch) {
  switch (ch) {
    case 'A':
      return 0;
    case 'C':
      return 1;
    case 'G':
      return 2;
    case 'T':
      return 3;
  }
}
void ins(string s, int len) {
  int p = 0;
  for (int i = 0; i < s.length(); ++i) {
    int idx = getId(s[i]);
    if (!son[p][idx]) son[p][idx] = ++tot;
    p = son[p][idx];
  }
  cnt[p] = len;
}
void getFail() {
  int head = 0, tail = 0;
  for (i = 0; i < 4; ++i) {
    if (son[0][i]) q[tail++] = son[0][i];
  }
  for (; head != tail;) {
    int u = q[head++];
    for (i = 0; i < 4; ++i) {
      if (!son[u][i])
        son[u][i] = son[fail[u]][i];
      else {
        fail[son[u][i]] = son[fail[u]][i];
        cnt[son[u][i]] = max(cnt[son[u][i]], cnt[fail[son[u][i]]]);
        q[tail++] = son[u][i];
      }
    }
  }
}
inline void up(int& a, int b) {
  a += b;
  if (a >= P) a -= P;
}
void solve() {
  int ans = 0;
  for (dp[0][0][0] = 1, i = 0; i < n; ++i)
    for (k = 0; k < 10; ++k)
      for (j = 0; j <= tot; ++j)
        for (l = 0; l < 4; ++l) {
          if (cnt[son[j][l]] >= k + 1)
            up(dp[i + 1][son[j][l]][0], dp[i][j][k]);
          else
            up(dp[i + 1][son[j][l]][k + 1], dp[i][j][k]);
        }
  for (i = 0; i <= tot; ++i) up(ans, dp[n][i][0]);
  printf("%d\n", ans);
}
int main() {
  scanf("%d%d", &n, &m);
  for (num = 0, i = 1; i <= m; ++i) {
    cin >> s;
    if (mp.find(s) == mp.end()) ins(s, (int)s.length()), mp[s] = 1;
  }
  getFail();
  solve();
  return 0;
}
