"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 md = 1000000009;
const int maxn = 12;
const int maxl = 1005;
const int Size = maxn * maxn;
char str[maxn][20];
int alp[300];
int nxt[Size][4];
int fail[Size];
int len[Size];
int rt, tol;
int dp[maxl][Size][11];
void add(int &a, int b) {
  (a += b);
  if (a >= md) a -= md;
}
int main() {
  alp['A'] = 0;
  alp['C'] = 1;
  alp['G'] = 2;
  alp['T'] = 3;
  rt = ++tol;
  len[rt] = 0;
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", str[i]);
    int p = rt, j;
    for (j = 0; str[i][j]; j++) {
      int c = alp[str[i][j]];
      if (!nxt[p][c]) {
        nxt[p][c] = ++tol;
      }
      p = nxt[p][c];
    }
    len[p] = max(len[p], j);
  }
  queue<int> q;
  for (int i = 0; i < 4; i++) {
    if (!nxt[rt][i])
      nxt[rt][i] = rt;
    else {
      fail[nxt[rt][i]] = rt;
      q.push(nxt[rt][i]);
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    len[u] = max(len[u], len[fail[u]]);
    for (int i = 0; i < 4; i++) {
      if (!nxt[u][i])
        nxt[u][i] = nxt[fail[u]][i];
      else {
        fail[nxt[u][i]] = nxt[fail[u]][i];
        q.push(nxt[u][i]);
      }
    }
  }
  dp[0][rt][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= tol; j++) {
      for (int d = 0; d <= 10; d++)
        if (dp[i - 1][j][d]) {
          for (int t = 0; t < 4; t++) {
            int k = nxt[j][t];
            if (k == rt) continue;
            if (len[k] > d)
              add(dp[i][k][0], dp[i - 1][j][d]);
            else if (d < 10)
              add(dp[i][k][d + 1], dp[i - 1][j][d]);
          }
        }
    }
  }
  int ans = 0;
  for (int i = 2; i <= tol; i++) add(ans, dp[n][i][0]);
  printf("%d\n", ans);
  return 0;
}
