"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 = 1000000009;
const string son = "ACGT";
char buf[20];
int T[110][4], fa[110], val[110], maxlen[110];
int fail[110][4];
long long f[1010][110][20];
string pre[110];
int tot;
string read() {
  scanf("%s", buf);
  return buf;
}
int key(char c) { return c == 'A' ? 0 : c == 'C' ? 1 : c == 'G' ? 2 : 3; }
int newnode(int f, int v) {
  fa[tot] = f, val[tot++] = v;
  return tot - 1;
}
void Insert(string &s) {
  int n = s.length(), now = 0;
  for (int i = 0; i < n; i++) {
    if (T[now][key(s[i])] == -1) T[now][key(s[i])] = newnode(now, key(s[i]));
    now = T[now][key(s[i])];
  }
  maxlen[now] = n;
}
void get_pre() {
  for (int i = 1; i < tot; i++) {
    pre[i] = pre[fa[i]] + son[val[i]];
  }
}
void get_maxlen() {
  int flag = 1;
  while (flag) {
    flag = 0;
    for (int i = 0; i < tot; i++) {
      for (int j = 0; j < tot; j++) {
        int n = pre[i].size(), m = pre[j].size();
        if (m > n && pre[j].substr(m - n, m) == pre[i]) {
          if (maxlen[j] < maxlen[i]) maxlen[j] = maxlen[i], flag = 1;
        }
      }
    }
  }
}
void get_fail() {
  for (int i = 0; i < tot; i++) {
    for (int q = 0; q < 4; q++) {
      string s = pre[i] + son[q];
      int maxj = 0, f = 0;
      for (int j = 1; j < tot; j++) {
        string t = pre[j];
        int n = t.size(), m = s.size();
        if (m >= n && s.substr(m - n, n) == t) {
          if (n > maxj) maxj = n, f = j;
        }
      }
      fail[i][q] = f;
    }
  }
}
int main() {
  memset(T, -1, sizeof(T));
  fa[tot] = -1, val[tot] = -1, pre[tot++] = "";
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    string s = read();
    Insert(s);
  }
  get_pre(), get_maxlen(), get_fail();
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < tot; j++) {
      for (int k = 0; k <= 10; k++) {
        if (f[i][j][k] == 0) continue;
        for (int q = 0; q < 4; q++) {
          int ii = i + 1, jj = fail[j][q], kk = maxlen[jj] >= k + 1 ? 0 : k + 1;
          f[ii][jj][kk] = (f[ii][jj][kk] + f[i][j][k]) % mod;
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < tot; i++) {
    ans = (ans + f[n][i][0]) % mod;
  }
  printf("%I64d\n", ans);
  return 0;
}
