"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 N = 10 * 20, P = 1000000009;
int n, m, top;
char str[200];
struct node {
  node *c[4], *fail;
  int len;
  int id;
};
int st, f[1020][N / 2 + 5][12];
node mem[N], *cur = mem, *q[N], *root, *nul = mem + 1;
int get(char x) {
  if (x == 'A') return 0;
  if (x == 'C') return 1;
  if (x == 'G') return 2;
  return 3;
}
node *newnode() {
  int i;
  cur++;
  cur->id = cur - mem;
  cur->len = 0;
  cur->fail = nul;
  for (i = 0; i < 4; i++) cur->c[i] = nul;
  top = cur->id;
  return cur;
}
void build_ac() {
  node *s, *v;
  int h, tail, i;
  q[h = tail = 1] = root;
  root->fail = root;
  while (h <= tail) {
    s = q[h++];
    for (i = 0; i < 4; i++) {
      if (s->c[i] == nul)
        s->c[i] = s->fail->c[i];
      else {
        q[++tail] = s->c[i];
        if (s != root) {
          v = s->fail;
          s->c[i]->fail = v->c[i];
          s->c[i]->len = max(s->c[i]->len, s->c[i]->fail->len);
        }
      }
    }
  }
  st = root->id;
}
void ins() {
  int idx, i;
  node *now = root;
  scanf("%s", str);
  for (i = 0; str[i]; i++) {
    idx = get(str[i]);
    if (now->c[idx] == nul) now->c[idx] = newnode();
    now = now->c[idx];
  }
  now->len = strlen(str);
}
void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
int main() {
  int i, j, idx, v, p;
  scanf("%d%d", &n, &m);
  nul = newnode();
  root = nul;
  while (m--) {
    ins();
  }
  build_ac();
  f[0][root->id][0] = 1;
  int k, t;
  for (i = 1; i <= n; i++)
    for (j = st; j <= top; j++)
      for (k = 0; k < 10; k++)
        if (f[i - 1][j][k])
          for (t = f[i - 1][j][k], p = 0; p < 4; p++) {
            v = mem[j].c[p]->id;
            if (k < mem[v].len)
              add(f[i][v][0], t);
            else
              add(f[i][v][k + 1], t);
          }
  int ans = 0;
  for (i = st; i <= top; i++) add(ans, f[n][i][0]);
  cout << ans;
}
