"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 add(int &a, int b) {
  a += b;
  if (a > mod) a -= mod;
  return a;
}
const int maxn = 1001, sigma = 26;
int len[maxn], link[maxn];
int to[maxn][sigma];
int sz = 1;
void add_str(string s) {
  int v = 0;
  for (auto c : s) {
    c -= 'A';
    if (!to[v][c]) to[v][c] = sz++;
    v = to[v][c];
  }
  len[v] = s.size();
}
void push_links() {
  int que[sz];
  int st = 0, fi = 1;
  que[0] = 0;
  while (st < fi) {
    int v = que[st++];
    int u = link[v];
    if (!len[v]) len[v] = len[u];
    for (int c = 0; c < sigma; c++)
      if (to[v][c]) {
        link[to[v][c]] = v ? to[u][c] : 0;
        que[fi++] = to[v][c];
      } else {
        to[v][c] = to[u][c];
      }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  int mlen = 0;
  while (m--) {
    string s;
    cin >> s;
    mlen = max(mlen, (int)s.size());
    add_str(s);
  }
  push_links();
  int dp[mlen][sz];
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (int z = 0; z < n; z++) {
    int dp1[mlen][sz];
    memset(dp1, 0, sizeof(dp1));
    for (int i = 0; i < sz; i++)
      for (int j = 0; j < mlen; j++)
        for (int c = 0; c < sigma; c++)
          if (len[to[i][c]] > j)
            add(dp1[0][to[i][c]], dp[j][i]);
          else if (j + 1 < mlen)
            add(dp1[j + 1][to[i][c]], dp[j][i]);
    memcpy(dp, dp1, sizeof(dp));
  }
  int ans = 0;
  for (int i = 0; i < sz; i++) add(ans, dp[0][i]);
  cout << ans << "\n";
}
