"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 sigma = 4;
struct Trie_node {
  Trie_node *next[sigma];
  Trie_node *fail;
  long long sign, len;
  void clear() {
    fail = 0;
    len = 0;
  }
};
Trie_node pool[1503000];
Trie_node *cur = pool;
Trie_node *newnode() {
  cur->clear();
  cur->sign = cur - pool + 1;
  return cur++;
}
vector<long long> last;
Trie_node *queue1[1010000];
struct Trie {
  Trie_node *root;
  Trie() { root = newnode(); }
  void insert(string &l) {
    Trie_node *found = root;
    for (auto &s : l) {
      if (found->next[s - 'a'] != NULL)
        found = found->next[s - 'a'];
      else {
        found->next[s - 'a'] = newnode();
        found = found->next[s - 'a'];
      }
    }
    found->len = l.size();
  }
  long long head, tail;
  void compute_fail() {
    head = 1;
    tail = 2;
    queue1[1] = root;
    while (head != tail) {
      Trie_node *part2 = NULL;
      Trie_node *father = queue1[head];
      head++;
      for (long long s = 0; s < sigma; ++s) {
        if (father->next[s] != NULL) {
          if (father == root)
            father->next[s]->fail = root;
          else {
            part2 = father->fail;
            while (part2 != NULL) {
              if (part2->next[s] != NULL) {
                father->next[s]->fail = part2->next[s];
                father->next[s]->len =
                    max(father->next[s]->len, father->next[s]->fail->len);
                break;
              }
              part2 = part2->fail;
            }
            if (part2 == NULL) father->next[s]->fail = root;
          }
          queue1[tail++] = (father->next[s]);
        }
      }
    }
  }
  void clear() { root = newnode(); }
} tr;
long long dp[1100][200][12];
long long mod = 1000000009;
map<char, char> trans;
string s[50];
Trie_node *pool2[2000];
long long dfs(long long len, long long cur, long long k) {
  if (k > 10) return 0;
  if (dp[len][cur][k] != -1) {
    return dp[len][cur][k];
  }
  if (len == 0 && k == 0) {
    return 1;
  } else if (len == 0)
    return 0;
  long long ans = 0;
  for (long long i = 0; i < 4; i++) {
    long long cur2 = cur;
    while (cur2 != 1 && pool2[cur2]->next[i] == NULL)
      cur2 = pool2[cur2]->fail->sign;
    if (pool2[cur2]->next[i] == NULL)
      continue;
    else {
      cur2 = pool2[cur2]->next[i]->sign;
      if (pool2[cur2]->len >= k + 1)
        ans += dfs(len - 1, cur2, 0);
      else
        ans += dfs(len - 1, cur2, k + 1);
      ans = ans % mod;
    }
  }
  dp[len][cur][k] = ans;
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, m;
  memset(dp, -1, sizeof(dp));
  trans['A'] = 'a';
  trans['T'] = 'b';
  trans['C'] = 'c';
  trans['G'] = 'd';
  cin >> n >> m;
  for (long long i = 1; i <= m; ++i) {
    cin >> s[i];
    if (s[i].size() > n) continue;
    for (auto &p : s[i]) p = trans[p];
    tr.insert(s[i]);
  }
  tr.compute_fail();
  Trie_node *kk = pool;
  while (kk != cur) {
    pool2[kk->sign] = kk;
    kk++;
  }
  cout << dfs(n, 1, 0) << endl;
  ;
  return 0;
}
