"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 MOD = 1e9 + 9;
int trans(char s) {
  if (s == 'A') return 0;
  if (s == 'C') return 1;
  if (s == 'G') return 2;
  if (s == 'T') return 3;
  return -1;
}
int n, m;
struct ACAutomaton {
  int tr[1005][4], tot = 1, fail[1005], dep[1005], epdep[1005], upper;
  bool ep[1005];
  void insert(char *str) {
    int len = strlen(str + 1), root = 1;
    upper = max(upper, len);
    for (int i = 1; i <= len; ++i) {
      int u = trans(str[i]);
      if (!tr[root][u]) tr[root][u] = ++tot;
      dep[tr[root][u]] = dep[root] + 1;
      root = tr[root][u];
    }
    ep[root] = true;
    epdep[root] = dep[root];
  }
  void buildFail() {
    ep[0] = true;
    for (int i = 0; i <= 3; ++i) tr[0][i] = 1;
    queue<int> Q;
    Q.push(1);
    while (!Q.empty()) {
      int now = Q.front();
      Q.pop();
      for (int i = 0; i <= 3; ++i) {
        if (tr[now][i]) {
          fail[tr[now][i]] = tr[fail[now]][i];
          if (ep[tr[fail[now]][i]]) ep[tr[now][i]] = ep[tr[fail[now]][i]];
          Q.push(tr[now][i]);
        } else
          tr[now][i] = tr[fail[now]][i];
      }
      epdep[now] = max(epdep[now], epdep[fail[now]]);
    }
  }
  int dp[1005][1005][15];
  int Dp() {
    dp[1][0][0] = 1;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j <= upper; ++j) {
        for (int k = 1; k <= tot; ++k) {
          for (int l = 0; l <= 3; ++l) {
            int v = tr[k][l];
            if (epdep[v] > j)
              (dp[v][i + 1][0] += dp[k][i][j]) %= MOD;
            else
              (dp[v][i + 1][j + 1] += dp[k][i][j]) %= MOD;
          }
        }
      }
    }
    int ans = 0;
    for (int i = 1; i <= tot; ++i) (ans += dp[i][n][0]) %= MOD;
    return ans;
  }
} AC;
char s[15];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; ++i) scanf("%s", s + 1), AC.insert(s);
  AC.buildFail();
  printf("%d", AC.Dp());
  return 0;
}
