"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 M = 12, L = 12, N = 1002;
const int Mod = 1e9 + 9;
char a[L], to[99];
int n;
int dp[2][M * L][L], ch[M * L][4], fail[M * L];
int len[N];
int ans = 0;
inline void mo(int& x) { x = (x >= Mod ? x - Mod : x); }
struct AC_automaton {
  int cnt;
  inline void init() {
    cnt = 0;
    to['A'] = 0, to['C'] = 1, to['G'] = 2, to['T'] = 3;
  }
  inline void insert() {
    int l = strlen(a + 1), now = 0;
    for (int i = 1; i <= l; i++) {
      if (!ch[now][to[a[i]]]) ch[now][to[a[i]]] = ++cnt;
      now = ch[now][to[a[i]]];
    }
    len[now] = max(len[now], l);
  }
  inline void build() {
    queue<int> Q;
    for (int i = 0; i < 4; 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[fail[x]]);
      for (int i = 0; i < 4; i++) {
        if (ch[x][i])
          fail[ch[x][i]] = ch[fail[x]][i], Q.push(ch[x][i]);
        else
          ch[x][i] = ch[fail[x]][i];
      }
    }
  }
  inline void work() {
    dp[0][0][0] = 1;
    for (int i = 1; i <= n; i++) {
      memset(dp[i & 1], 0, sizeof(dp[i & 1]));
      for (int j = 0; j <= cnt; j++)
        for (int k = 0; k <= 10; k++) {
          if (dp[!(i & 1)][j][k]) {
            for (int t = 0; t < 4; t++) {
              if (len[ch[j][t]] > k)
                mo(dp[i & 1][ch[j][t]][0] += dp[!(i & 1)][j][k]);
              else
                mo(dp[i & 1][ch[j][t]][k + 1] += dp[!(i & 1)][j][k]);
            }
          }
        }
    }
    for (int j = 0; j <= cnt; j++) mo(ans += dp[n & 1][j][0]);
  }
} T;
int main() {
  int m;
  scanf("%d%d", &n, &m);
  T.init();
  for (int i = 1; i <= m; i++) {
    scanf("%s", a + 1);
    T.insert();
  }
  T.build();
  T.work();
  printf("%d", ans);
  return 0;
}
