"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 maxn = 500010;
const int maxm = 1000010;
const long long mod = 1e9 + 9;
struct AcTrie {
  int next[maxn][4], fail[maxn];
  int end[maxn];
  int root, tot;
  int newnode() {
    for (int i = 0; i < 4; ++i) next[tot][i] = -1;
    end[tot++] = 0;
    return tot - 1;
  }
  int flx(char c) {
    if (c == 'A')
      return 0;
    else if (c == 'C')
      return 1;
    else if (c == 'G')
      return 2;
    else
      return 3;
  }
  void init() {
    tot = 0;
    root = newnode();
  }
  void insert(char *buf) {
    int len = strlen(buf), now = root;
    for (int i = 0; i < len; ++i) {
      if (next[now][flx(buf[i])] == -1) next[now][flx(buf[i])] = newnode();
      now = next[now][flx(buf[i])];
    }
    end[now] = len;
  }
  void build() {
    queue<int> Q;
    fail[root] = root;
    for (int i = 0; i < 4; ++i) {
      if (next[root][i] == -1)
        next[root][i] = root;
      else {
        fail[next[root][i]] = root;
        Q.push(next[root][i]);
      }
    }
    while (!Q.empty()) {
      int now = Q.front();
      Q.pop();
      for (int i = 0; i < 4; ++i) {
        if (next[now][i] == -1)
          next[now][i] = next[fail[now]][i];
        else {
          fail[next[now][i]] = next[fail[now]][i];
          end[next[now][i]] = max(end[next[now][i]], end[fail[next[now][i]]]);
          Q.push(next[now][i]);
        }
      }
    }
  }
  void debug() {
    for (int i = 0; i < tot; ++i) {
      printf("%d %d\n[", end[i], fail[i]);
      for (int j = 0; j < 4; ++j) printf("%d ", next[i][j]);
      printf("]\n");
    }
  }
};
AcTrie ac;
int n, m;
char s[20];
long long dp[1010][110][15];
void solve() {
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < ac.tot; ++j) {
      for (int k = 0; k < 10; ++k) {
        for (int t = 0; t < 4; ++t) {
          if (ac.next[j][t] == 0) continue;
          if (ac.end[ac.next[j][t]] >= k + 1) {
            dp[i + 1][ac.next[j][t]][0] += dp[i][j][k];
            dp[i + 1][ac.next[j][t]][0] %= mod;
          } else {
            dp[i + 1][ac.next[j][t]][k + 1] += dp[i][j][k];
            dp[i + 1][ac.next[j][t]][k + 1] %= mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < ac.tot; ++i) {
    ans += dp[n][i][0];
    ans %= mod;
  }
  printf("%lld\n", ans);
}
int main() {
  scanf("%d %d", &n, &m);
  ac.init();
  while (m--) {
    scanf("%s", s);
    ac.insert(s);
  }
  ac.build();
  solve();
}
