"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 = 10 * 10 + 5;
const int CHAR_SET = 4;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 9;
long long dp[1005][N][12];
class trie_2 {
 private:
  vector<vector<int> > ch;
  vector<int> len, fail, q;
  int n, cnt, front, rear;
  int get_key(char c) const {
    if (c == 'A')
      return 0;
    else if (c == 'T')
      return 1;
    else if (c == 'C')
      return 2;
    else
      return 3;
  }

 public:
  explicit trie_2(int _n = N)
      : ch(_n, vector<int>(CHAR_SET)), len(_n), fail(_n), q(_n), n(1), cnt(1) {}
  void insert(const char* str, int l) {
    int u = 1;
    while (true) {
      if (*str == 0) {
        len[u] = l;
        break;
      }
      int v = get_key(*str);
      if (!ch[u][v]) ch[u][v] = ++cnt;
      u = ch[u][v];
      ++str;
    }
  }
  void build_trie_2() {
    front = 0;
    rear = 0;
    for (int i = 0; i < CHAR_SET; i++)
      if (ch[1][i]) {
        fail[ch[1][i]] = 1;
        q[rear++] = ch[1][i];
      } else
        ch[1][i] = 1;
    while (front != rear) {
      int u = q[front++];
      if (!len[u]) len[u] = len[fail[u]];
      for (int i = 0; i < CHAR_SET; i++)
        if (ch[u][i]) {
          fail[ch[u][i]] = ch[fail[u]][i];
          q[rear++] = ch[u][i];
        } else
          ch[u][i] = ch[fail[u]][i];
    }
  }
  long long solve(int n) {
    memset((dp), (0), sizeof(dp));
    dp[0][1][0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= cnt; j++) {
        for (int k = 0; k <= 10; k++) {
          if (!dp[i - 1][j][k]) continue;
          for (int c = 0; c < CHAR_SET; c++) {
            int son = ch[j][c];
            if (len[son] >= k + 1)
              dp[i][son][0] = (dp[i][son][0] + dp[i - 1][j][k]) % MOD;
            else
              dp[i][son][k + 1] = (dp[i][son][k + 1] + dp[i - 1][j][k]) % MOD;
          }
        }
      }
    }
    long long ans = 0;
    for (int i = 1; i <= cnt; i++) ans = (ans + dp[n][i][0]) % MOD;
    return ans;
  }
} trie;
char s[15];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%s", s);
    trie.insert(s, strlen(s));
  }
  trie.build_trie_2();
  printf("%lld\n", trie.solve(n));
  return 0;
}
