"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 maxn = 110, maxl = 1010, inf = 1000000009;
int n, m, t;
int q[maxn];
int p[maxn][26], en[maxn], last[maxn], dep[maxn];
int f[maxl][12][maxn];
void add(string s) {
  int start = 0, len = s.size();
  for (int i = 0; i < len; i++)
    if (p[start][s[i] - 'A'] == 0) {
      start = p[start][s[i] - 'A'] = ++t;
      dep[start] = i + 1;
    } else
      start = p[start][s[i] - 'A'];
  en[start] = len;
}
void find(int x) {
  int now = x;
  while (now) {
    if (en[now]) {
      en[x] = en[now];
      break;
    }
    now = last[now];
  }
}
void build() {
  int head = 0, tail = 0;
  for (int i = 0; i < 26; i++)
    if (p[0][i]) q[++tail] = p[0][i];
  for (; head < tail;) {
    int u = q[++head];
    for (int i = 0; i < 26; i++)
      if (p[u][i]) {
        q[++tail] = p[u][i];
        last[p[u][i]] = p[last[u]][i];
      } else
        p[u][i] = p[last[u]][i];
  }
  for (int i = 1; i <= t; i++) find(i);
}
void update(int &x, int y) {
  x += y;
  if (x >= inf) x -= inf;
}
void dp() {
  memset(f, 0, sizeof(f));
  f[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= 10; j++)
      for (int k = 0; k <= t; k++)
        if (f[i][j][k])
          for (int nc = 0; nc < 26; nc++) {
            int u = p[k][nc];
            if (en[u] >= j + 1)
              update(f[i + 1][0][u], f[i][j][k]);
            else
              update(f[i + 1][j + 1][u], f[i][j][k]);
          }
  int ans = 0;
  for (int i = 0; i <= t; i++) update(ans, f[n][0][i]);
  cout << ans << endl;
}
int main() {
  t = 0;
  memset(p, 0, sizeof(p));
  memset(en, 0, sizeof(en));
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    string s;
    cin >> s;
    add(s);
  }
  build();
  dp();
  return 0;
}
