"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 kMaxStates = 1 << 7;
const int kMaxN = 1 << 10;
const int kAlph = 4;
int t[kMaxStates][kAlph], T, len[kMaxStates], out[kMaxStates];
int q[kMaxStates], jump[kMaxStates];
inline int index(char ch) {
  static const string bio = "ACGT";
  return bio.find(ch);
}
void add(const string &s) {
  int cur = 1;
  for (int i = 0; i < s.size(); ++i) {
    int code = index(s[i]);
    if (!t[cur][code]) {
      len[t[cur][code] = ++T] = 1 + len[cur];
    }
    cur = t[cur][code];
  }
  out[cur] = s.size();
}
void build() {
  int head = 0, rear = 0;
  for (int i = 0; i < kAlph; ++i)
    if (t[1][i]) {
      jump[t[1][i]] = 1;
      q[rear++] = t[1][i];
    }
  for (int i = 0; i < kAlph; ++i)
    if (!t[1][i]) t[1][i] = 1;
  while (head < rear) {
    int r = q[head++];
    for (int i = 0; i < kAlph; ++i)
      if (t[r][i]) {
        int u = t[r][i];
        q[rear++] = u;
        int v = jump[r];
        while (!t[v][i]) v = jump[v];
        jump[u] = t[v][i];
        out[u] = max(out[u], out[jump[u]]);
      }
  }
}
int dp[kMaxN][12][kMaxStates], n, maxL;
const int mod = 1E9 + 9;
int rec(int cur, int left, int node) {
  if (cur == n) return left == 0;
  if (left > maxL) return 0;
  int &res = dp[cur][left][node];
  if (res == -1) {
    res = 0;
    for (int i = 0; i < kAlph; ++i) {
      if (t[node][i] && t[node][i] != node) {
        int p = t[node][i];
        if (len[p] >= 1 + left) {
          res += rec(cur + 1, out[p] >= left + 1 ? 0 : left + 1, p);
          if (res >= mod) res -= mod;
        }
      } else {
        int v = node;
        while (!t[v][i]) v = jump[v];
        if (t[v][i] && t[v][i] != v) {
          int p = t[v][i];
          if (len[p] >= 1 + left) {
            res += rec(cur + 1, out[p] >= left + 1 ? 0 : left + 1, p);
            if (res >= mod) res -= mod;
          }
        }
      }
    }
  }
  return res;
}
int main() {
  T = 1;
  int m;
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    string s;
    cin >> s;
    add(s);
    maxL = max(maxL, (int)s.size());
  }
  build();
  memset(dp, -1, sizeof dp);
  const int res = rec(0, 0, 1);
  cout << res << endl;
  return 0;
}
