"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 mod = 1e9 + 9;
const int MAXN = 1005, MAXM = 15, MAXL = 105;
template <typename _T>
void read(_T &x) {
  x = 0;
  char s = getchar();
  int f = 1;
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s - '0'), s = getchar();
  }
  x *= f;
}
template <typename _T>
void write(_T x) {
  if (x < 0) {
    putchar('-');
    x = (~x) + 1;
  }
  if (9 < x) {
    write(x / 10);
  }
  putchar(x % 10 + '0');
}
template <typename _T>
_T MIN(const _T a, const _T b) {
  return a < b ? a : b;
}
template <typename _T>
_T MAX(const _T a, const _T b) {
  return a > b ? a : b;
}
int f[MAXN][MAXL][MAXM];
int ch[MAXL][4], fail[MAXL], q[MAXL], len[MAXL];
char S[MAXM];
int N, M, cnt, mx;
void add(int &x, const int v) { x = (x + v >= mod ? x + v - mod : x + v); }
int getID(const char C) {
  if (C == 'A') return 0;
  if (C == 'T') return 1;
  if (C == 'C') return 2;
  if (C == 'G') return 3;
}
void insert(char *T) {
  int p = 0, id;
  for (int i = 1; T[i]; i++) {
    id = getID(T[i]);
    if (!ch[p][id]) ch[p][id] = ++cnt;
    p = ch[p][id];
  }
  mx = MAX(mx, len[p] = MAX(len[p], (int)strlen(T + 1)));
}
void init() {
  int h = 1, t = 0, u, v;
  for (int i = 0; i < 4; i++)
    if (ch[0][i]) q[++t] = ch[0][i];
  while (h <= t) {
    u = q[h++];
    for (int i = 0; i < 4; i++) {
      if (v = ch[u][i])
        fail[v] = ch[fail[u]][i], q[++t] = v;
      else
        ch[u][i] = ch[fail[u]][i];
    }
    len[u] = MAX(len[u], len[fail[u]]);
  }
}
int main() {
  read(N), read(M);
  for (int i = 1; i <= M; i++) scanf("%s", S + 1), insert(S);
  init();
  f[0][0][0] = 1;
  for (int i = 0; i < N; i++)
    for (int p = 0; p <= cnt; p++)
      for (int j = 0; j <= mx; j++)
        if (f[i][p][j])
          for (int c = 0; c < 4; c++)
            add(f[i + 1][ch[p][c]][(len[ch[p][c]] > j ? 0 : j + (j != mx))],
                f[i][p][j]);
  int ans = 0;
  for (int p = 0; p <= cnt; p++) add(ans, f[N][p][0]);
  write(ans), putchar('\n');
  return 0;
}
