"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 long long mod = 1000000009;
int n, m, fail[1005];
int nxt[1005][26], sum[1005], ndsum;
long long f[1005][1005][11];
char a[15];
bool is_end[1005];
void ins(char a[]) {
  int len = strlen(a), now = 0;
  for (int i = 0; i < len; ++i) {
    int c = a[i] - 'A';
    if (!nxt[now][c]) nxt[now][c] = ++ndsum;
    now = nxt[now][c];
  }
  sum[now] = len;
  is_end[now] = 1;
}
void get_fail() {
  queue<int> q;
  for (int i = 0; i < 26; ++i)
    if (nxt[0][i]) q.push(nxt[0][i]);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int j = 0; j < 26; ++j) {
      if (nxt[u][j]) {
        q.push(nxt[u][j]);
        fail[nxt[u][j]] = nxt[fail[u]][j];
        sum[nxt[u][j]] = ((sum[nxt[u][j]]) > (sum[fail[nxt[u][j]]])
                              ? (sum[nxt[u][j]])
                              : (sum[fail[nxt[u][j]]]));
        is_end[nxt[u][j]] |= is_end[nxt[fail[u]][j]];
      } else
        nxt[u][j] = nxt[fail[u]][j];
    }
  }
}
void dp() {
  f[0][0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= ndsum; ++j)
      for (int k = 0; k < 10; ++k)
        if (f[i - 1][j][k])
          for (int c = 0; c < 26; ++c) {
            int now = nxt[j][c];
            if (sum[now] >= k + 1)
              f[i][now][0] = (f[i][now][0] + f[i - 1][j][k]) % mod;
            else
              f[i][now][k + 1] = (f[i][now][k + 1] + f[i - 1][j][k]) % mod;
          }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    scanf("%s", a);
    ins(a);
  }
  get_fail();
  dp();
  long long ans = 0;
  for (int i = 0; i <= ndsum; ++i)
    if (is_end[i]) ans = (ans + f[n][i][0]) % mod;
  cout << ans << endl;
  return 0;
}
