"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>
const int MAXN = 1005;
const int MOD = 1e9 + 9;
int N, M, RFL[30];
char seq[15];
namespace AC {
int sz;
void insert(const char* st);
void constructer();
int calc();
}  // namespace AC
int main() {
  char st[] = "ACGT";
  for (int i = 0; st[i]; i++) RFL[st[i] - 'A'] = i;
  AC::sz = 1;
  scanf("%d%d", &N, &M);
  for (int i = 0; i < M; i++) {
    scanf("%s", seq);
    AC::insert(seq);
  }
  AC::constructer();
  printf("%d\n", AC::calc());
  return 0;
}
namespace AC {
const int LEN = 105;
int T[LEN][4], mxlen[LEN], fail[LEN], DP[MAXN][LEN][11];
void insert(const char* st) {
  int i, rt, ch;
  for (i = rt = 0, ch; st[i]; i++) {
    if (!T[rt][ch = RFL[st[i] - 'A']]) T[rt][ch] = sz++;
    rt = T[rt][ch];
  }
  mxlen[rt] = i;
}
void constructer() {
  std::queue<int> que;
  for (int i = 0; i < 4; i++)
    if (T[0][i]) que.push(T[0][i]);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = 0; i < 4; i++) {
      int& v = T[u][i];
      if (v) {
        fail[v] = T[fail[u]][i];
        mxlen[v] = std::max(mxlen[v], mxlen[fail[v]]);
        que.push(v);
      } else
        v = T[fail[u]][i];
    }
  }
}
inline void add(int& x, int y) { (x += y) >= MOD ? x -= MOD : 0; }
int calc() {
  DP[0][0][0] = 1;
  for (int i = 0; i < N; i++)
    for (int u = 0; u < sz; u++)
      for (int j = 0; j < 10; j++) {
        if (!DP[i][u][j]) continue;
        for (int ch = 0; ch < 4; ch++) {
          int v = T[u][ch];
          if (mxlen[v] >= j + 1)
            add(DP[i + 1][v][0], DP[i][u][j]);
          else
            add(DP[i + 1][v][j + 1], DP[i][u][j]);
        }
      }
  int ret = 0;
  for (int i = 0; i < sz; i++) add(ret, DP[N][i][0]);
  return ret;
}
}  // namespace AC
