"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;
int n, m;
const int MOD = 1e9 + 9;
int idx(char c) {
  if (c == 'A') return 0;
  if (c == 'T') return 1;
  if (c == 'G') return 2;
  if (c == 'C') return 3;
  assert(0);
  return -1;
}
const int _ = 120;
int tag[_];
namespace ACAM {
const int _ = 120;
int tr[_][4], fail[_], tot = 0;
void insert(char* S) {
  int u = 0, len = 0;
  for (int i = 0; S[i]; i++) {
    int ex = idx(S[i]);
    if (!tr[u][ex]) tr[u][ex] = ++tot;
    u = tr[u][ex];
    len++;
  }
  assert(tag[u] <= len);
  tag[u] = len;
}
void build() {
  queue<int> Q;
  for (int i = 0; i < 4; i++)
    if (tr[0][i]) Q.push(tr[0][i]);
  while (!Q.empty()) {
    int now = Q.front();
    Q.pop();
    for (int i = 0; i < 4; i++) {
      if (tr[now][i]) {
        fail[tr[now][i]] = tr[fail[now]][i], Q.push(tr[now][i]);
      } else
        tr[now][i] = tr[fail[now]][i];
    }
    tag[now] = max(tag[now], tag[fail[now]]);
  }
}
}  // namespace ACAM
int dp[1010][120][12];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    static char S[_];
    scanf("%s", S);
    ACAM::insert(S);
  }
  ACAM::build();
  int KK = ACAM::tot;
  dp[0][0][0] = 1;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= KK; j++) {
      for (int k = 0; k <= 10; k++) {
        int now = dp[i][j][k];
        for (int t = 0; t < 4; t++) {
          int ex = ACAM::tr[j][t];
          if (k < tag[ex])
            (dp[i + 1][ex][0] += now) %= MOD;
          else
            (dp[i + 1][ex][k + 1] += now) %= MOD;
        }
      }
    }
  }
  int ret = 0;
  for (int i = 1; i <= KK; i++) {
    (ret += dp[n][i][0]) %= MOD;
  }
  printf("%d\n", ret);
  return 0;
}
