"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 MOD = 1E9 + 9;
int n, m, ans, cntNode(1),
    root(1), f[200], dp[1005][105][13], vis[1005][105][13];
string s;
struct Trie {
  int son[4], fail, len;
  Trie() {
    for (int i = 0; i < 4; i++) son[i] = 0;
    fail = len = 0;
  }
} tr[105];
void add(int& a, int b) { a = (a + b) % MOD; }
void insert(string& s) {
  int cur = root;
  for (int i = 0; i < s.size(); i++) {
    if (!tr[cur].son[f[s[i]]]) tr[cur].son[f[s[i]]] = ++cntNode;
    cur = tr[cur].son[f[s[i]]];
  }
  tr[cur].len = s.size();
}
void getFail() {
  queue<int> Q;
  Q.push(root);
  tr[root].fail = root;
  while (!Q.empty()) {
    int now = Q.front();
    Q.pop();
    tr[now].len = max(tr[now].len, tr[tr[now].fail].len);
    for (int i = 0; i < 4; i++) {
      if (tr[now].son[i]) {
        tr[tr[now].son[i]].fail = now != root ? tr[tr[now].fail].son[i] : root;
        Q.push(tr[now].son[i]);
      } else
        tr[now].son[i] =
            tr[tr[now].fail].son[i] ? tr[tr[now].fail].son[i] : root;
    }
  }
}
void DP() {
  queue<tuple<int, int, int> > Q;
  dp[0][1][0] = 1;
  Q.push(make_tuple(0, 1, 0));
  while (!Q.empty()) {
    int len = get<0>(Q.front());
    int node = get<1>(Q.front());
    int last = get<2>(Q.front());
    Q.pop();
    if (len == n) {
      if (!last) add(ans, dp[len][node][last]);
      continue;
    }
    vis[len][node][last] = 0;
    for (int i = 0; i < 4; i++) {
      int to = tr[node].son[i];
      int tmp = tr[to].len > last ? 0 : last + 1;
      if (tmp > 10) continue;
      add(dp[len + 1][to][tmp], dp[len][node][last]);
      if (!vis[len + 1][to][tmp]) {
        vis[len + 1][to][tmp] = 1;
        Q.push(make_tuple(len + 1, to, tmp));
      }
    }
  }
  return;
}
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  f['A'] = 0, f['T'] = 1, f['C'] = 2, f['G'] = 3;
  for (int i = 1; i <= m; i++) cin >> s, insert(s);
  getFail();
  DP();
  cout << ans;
  return 0;
}
