"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 N = 1010, M = 100005, oo = 0x3f3f3f3f, mod = 1000000009;
int idn[300], n, m, pt = 1, dp[N][110][12], siz, ans = 0;
char in[12];
struct node {
  node *son[4], *fail;
  int val;
} pl[M], *sz = pl, *root;
node* newnode() {
  node* rec = sz++;
  memset(rec->son, 0, sizeof(rec->son));
  rec->fail = 0;
  rec->val = 0;
  return rec;
}
void init() {
  sz = pl;
  root = newnode();
}
void insert(char* s) {
  int len = strlen(s);
  node* p = root;
  for (int i = 0; i < len; i++) {
    int t = idn[s[i]];
    if (p->son[t] == NULL) p->son[t] = newnode();
    p = p->son[t];
  }
  p->val = len;
}
void build() {
  static queue<node*> q;
  root->fail = root;
  for (int i = 0; i < 4; i++) {
    if (root->son[i])
      root->son[i]->fail = root, q.push(root->son[i]);
    else
      root->son[i] = root;
    if (root->son[i]->fail->val)
      root->son[i]->val = max(root->son[i]->val, root->son[i]->fail->val);
  }
  while (!q.empty()) {
    node* p = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      if (p->son[i]) {
        p->son[i]->fail = p->fail->son[i];
        q.push(p->son[i]);
        if (p->son[i]->fail->val)
          p->son[i]->val = max(p->son[i]->val, p->son[i]->fail->val);
      } else
        p->son[i] = p->fail->son[i];
    }
  }
}
int main() {
  idn['A'] = 0;
  idn['G'] = 1;
  idn['C'] = 2;
  idn['T'] = 3;
  init();
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) scanf("%s", in), insert(in);
  build();
  siz = sz - pl;
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < siz; j++)
      for (int k = 0; k <= 10; k++)
        if (dp[i][j][k]) {
          for (int t = 0; t < 4; t++) {
            node* p = (pl + j)->son[t];
            if (p->val >= k + 1)
              (dp[i + 1][p - pl][0] += dp[i][j][k]) %= mod;
            else
              (dp[i + 1][p - pl][k + 1] += dp[i][j][k]) %= mod;
          }
        }
  for (int t = 0; t < siz; t++) (ans += dp[n][t][0]) %= mod;
  cout << ans;
  return 0;
}
