"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 NODE = 11 * 11;
const int MOD = 1000000009LL;
int T[322];
int edge[NODE][4];
int pre[NODE];
int state[NODE];
queue<int> que;
int ecnt;
void clr() {
  memset(edge, -1, sizeof(edge));
  memset(pre, -1, sizeof(pre));
  pre[0] = 0;
  ecnt = 0;
}
void add(char *s) {
  int pos = 0;
  int len = 0;
  for (; *s; s++) {
    len++;
    if (edge[pos][T[*s]] == -1) {
      edge[pos][T[*s]] = ++ecnt;
      state[ecnt] = 0;
    }
    pos = edge[pos][T[*s]];
  }
  state[pos] = len;
}
bool vis[NODE] = {0};
void build() {
  for (int c = 0; c < 4; c++)
    if (edge[0][c] == -1)
      edge[0][c] = 0;
    else {
      que.push(edge[0][c]);
      pre[edge[0][c]] = 0;
    }
  while (!que.empty()) {
    int ip = que.front();
    que.pop();
    for (int c = 0; c < 4; c++) {
      if (edge[ip][c] == -1)
        edge[ip][c] = edge[pre[ip]][c];
      else {
        pre[edge[ip][c]] = edge[pre[ip]][c];
        state[edge[ip][c]] = max(state[edge[pre[ip]][c]], state[edge[ip][c]]);
        que.push(edge[ip][c]);
      }
    }
  }
}
int dp[NODE][1001][11];
int calc(int p, int n, int wr) {
  if (wr > 10) return 0;
  if (n == 0 && wr == 0) return 1;
  if (n == 0) return 0;
  if (dp[p][n][wr] != -1) return dp[p][n][wr];
  int &ans = dp[p][n][wr];
  ans = 0;
  for (int c = 0; c < 4; c++) {
    int np = edge[p][c];
    int nwr = wr + 1;
    if (state[np] >= nwr) nwr = 0;
    ans = (ans + calc(np, n - 1, nwr)) % MOD;
  }
  return ans;
}
int main() {
  int n, m;
  T['A'] = 0;
  T['G'] = 1;
  T['C'] = 2;
  T['T'] = 3;
  scanf("%d%d", &n, &m);
  getchar();
  char ts[100];
  clr();
  for (int i = 0; i < m; i++) {
    gets(ts);
    add(ts);
  }
  build();
  memset(dp, -1, sizeof(dp));
  int ans = calc(0, n, 0);
  printf("%d\n", ans);
}
