"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 MODULE = (int)1e9 + 9;
struct TNODE {
  TNODE* Child[4];
  TNODE* Fail;
  int len;
};
TNODE TPool[111];
int TPTop = 0;
TNODE* Root = NULL;
int chrid[256] = {0};
int trie_insert(char* text, int len) {
  TNODE* x = Root;
  while (*text) {
    int ch = chrid[*text];
    if (!x->Child[ch]) {
      x->Child[ch] = &TPool[TPTop++];
    }
    x = x->Child[ch];
    text++;
  }
  x->len = max(x->len, len);
  return 0;
}
int Aho_Corasick() {
  queue<TNODE*> q;
  q.push(Root);
  while (!q.empty()) {
    TNODE* x = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      if (x->Child[i]) {
        x->Child[i]->Fail = x->Fail ? x->Fail->Child[i] : Root;
        x->Child[i]->len = max(x->Child[i]->len, x->Child[i]->Fail->len);
        q.push(x->Child[i]);
      } else
        x->Child[i] = x->Fail ? x->Fail->Child[i] : Root;
    }
  }
  return 0;
}
int F[2][111][11];
int main(void) {
  chrid['A'] = 0;
  chrid['T'] = 1;
  chrid['C'] = 2;
  chrid['G'] = 3;
  Root = &TPool[TPTop++];
  int mxlen = 0;
  int n = 0;
  int m = 0;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < m; i++) {
    char dna[233] = {0};
    scanf("%s", dna);
    int len = strlen(dna);
    trie_insert(dna, len);
    mxlen = max(mxlen, len);
  }
  Aho_Corasick();
  F[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    int ci = i & 1;
    int ni = ci ^ 1;
    memset(F[ni], 0, sizeof(F[ni]));
    for (int j = 0; j < TPTop; j++) {
      for (int k = 0; k <= mxlen; k++) {
        if (!F[ci][j][k]) continue;
        for (int ch = 0; ch < 4; ch++) {
          int tar = TPool[j].Child[ch] - TPool;
          if (TPool[tar].len >= k + 1)
            F[ni][tar][0] = (F[ni][tar][0] + F[ci][j][k]) % MODULE;
          else if (k + 1 <= mxlen)
            F[ni][tar][k + 1] = (F[ni][tar][k + 1] + F[ci][j][k]) % MODULE;
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < TPTop; i++) ans = (ans + F[n & 1][i][0]) % MODULE;
  printf("%d\n", ans);
  while (getchar() != EOF)
    ;
  return 0;
}
