"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>
int tr[102][4], fail[102], sum[102], len[102], cnt;
int dp[102][1002][12];
const int p = 1e9 + 9;
int t[128];
inline int max(int x, int y) { return x > y ? x : y; }
inline void insert(int _len, char *s) {
  int u = 0;
  for (int i = 1; i <= _len; i++) {
    int c = t[s[i]];
    if (!tr[u][c]) tr[u][c] = ++cnt;
    u = tr[u][c];
  }
  sum[u]++, len[u] = max(len[u], _len);
}
struct Queue {
  int q[102], head, tail;
  inline void clear() { head = 1, tail = 0; }
  inline void push(int x) { q[++tail] = x; }
  inline void pop() { head++; }
  inline bool empty() { return head > tail; }
  inline int front() { return q[head]; }
} q;
int node[102];
inline void ACAM() {
  q.clear();
  for (int i = 0; i < 4; i++)
    if (tr[0][i]) q.push(tr[0][i]);
  for (int i = 1; !q.empty(); i++) {
    int u = q.front();
    q.pop();
    node[i] = u;
    sum[u] += sum[fail[u]];
    len[u] = max(len[u], len[fail[u]]);
    for (int i = 0; i < 4; i++)
      if (tr[u][i])
        fail[tr[u][i]] = tr[fail[u]][i], q.push(tr[u][i]);
      else
        tr[u][i] = tr[fail[u]][i];
  }
}
int n, m;
char s[102];
int main() {
  t['A'] = 0, t['T'] = 1, t['C'] = 2, t['G'] = 3;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%s", s + 1), insert(strlen(s + 1), s);
  ACAM();
  dp[0][0][0] = 1;
  for (int l = 0; l < n; l++)
    for (int _i = 0, u = 0; _i <= cnt; _i++, u = node[_i])
      for (int j = 0; j < 4; j++)
        for (int i = 0; i <= 10; i++)
          if (len[tr[u][j]] >= i + 1)
            (dp[tr[u][j]][l + 1][0] += dp[u][l][i]) %= p;
          else
            (dp[tr[u][j]][l + 1][i + 1] += dp[u][l][i]) %= p;
  int ans = 0;
  for (int i = 0; i <= cnt; i++) ans = (ans + dp[i][n][0]) % p;
  printf("%d", ans);
  return 0;
}
