"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 cc = 1000000009ll;
struct trie {
  trie* next[5];
  trie* fail;
  int l;
  int f;
} * root, *alloc, sav[1005];
char s[30];
queue<trie*> q;
long long dp[2005][205][22];
int n, m;
int cnt;
int ch(char c) {
  if (c == 'A') return 0;
  if (c == 'C') return 1;
  if (c == 'G') return 2;
  if (c == 'T') return 3;
}
void ins(char* s) {
  int l = strlen(s);
  trie* p = root;
  for (int i = 0; i < l; i++) {
    int w = ch(s[i]);
    if (p->next[w] == NULL) {
      memset(alloc, 0, sizeof(trie));
      p->next[w] = alloc;
      p->next[w]->f = cnt;
      alloc++;
      cnt++;
    }
    p = p->next[w];
  }
  p->l = strlen(s);
}
void build() {
  while (!q.empty()) q.pop();
  root->fail = NULL;
  q.push(root);
  while (!q.empty()) {
    trie* cur = q.front();
    q.pop();
    for (int i = 0; i < 4; i++)
      if (cur->next[i] != NULL) {
        if (cur == root)
          cur->next[i]->fail = root;
        else {
          trie* p = cur->fail;
          cur->next[i]->fail = p->next[i];
          if (cur->next[i]->fail == NULL) cur->next[i]->fail = root;
          cur->next[i]->l = max(cur->next[i]->l, cur->next[i]->fail->l);
        }
        q.push(cur->next[i]);
      } else {
        if (cur == root)
          cur->next[i] = root;
        else
          cur->next[i] = cur->fail->next[i];
      }
  }
}
int main() {
  while (scanf("%d%d", &n, &m) == 2) {
    alloc = sav;
    memset(alloc, 0, sizeof(trie));
    root = alloc;
    root->f = 0;
    alloc++;
    cnt = 1;
    for (int i = 0; i < m; i++) {
      scanf("%s", s);
      ins(s);
    }
    build();
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < cnt; j++)
        for (int k = 0; k <= 10; k++)
          if (dp[i][j][k]) {
            for (int u = 0; u < 4; u++)
              if (sav[j].next[u] != NULL) {
                int nex = sav[j].next[u]->f;
                int tl = sav[j].next[u]->l;
                if (tl > k) {
                  dp[i + 1][nex][0] += dp[i][j][k];
                  dp[i + 1][nex][0] %= cc;
                } else {
                  if (k + 1 > 10) continue;
                  dp[i + 1][nex][k + 1] += dp[i][j][k];
                  dp[i + 1][nex][k + 1] %= cc;
                }
              }
          }
    long long ans = 0;
    for (int j = 0; j < cnt; j++) ans = (ans + dp[n][j][0]) % cc;
    printf("%I64d\n", ans);
  }
}
