"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;
struct node {
  int p[4], fail, d, B;
  bool b;
} t[105];
int n, m, l[15], w, dp[1005][105][15], out;
char s[15][15];
void build() {
  for (int i = 0; i < 4; i++) {
    t[w].p[i] = 0;
  }
  t[w].fail = -1;
  t[w].d = 0;
  t[w].B = 0;
  t[w].b = false;
  w++;
}
char ref(char c) {
  if (c == 'A') return 0;
  if (c == 'C') return 1;
  if (c == 'G') return 2;
  if (c == 'T') return 3;
}
void ins(int z) {
  int c = 0;
  for (int i = 0; i < l[z]; i++) {
    s[z][i] = ref(s[z][i]);
    if (t[c].p[s[z][i]] == 0) {
      t[c].p[s[z][i]] = w;
      build();
    }
    c = t[c].p[s[z][i]];
  }
  t[c].b = true;
}
void ac() {
  int q[105], rq, wq;
  rq = wq = 0;
  q[wq++] = 0;
  while (rq != wq) {
    int c = q[rq];
    for (int i = 0; i < 4; i++) {
      if (t[c].p[i]) {
        int temp = t[c].fail;
        while (temp != -1) {
          if (t[temp].p[i]) {
            temp = t[temp].p[i];
            break;
          }
          temp = t[temp].fail;
        }
        if (temp == -1) temp = 0;
        t[t[c].p[i]].fail = temp;
        t[t[c].p[i]].d = t[c].d + 1;
        if (t[t[c].p[i]].b)
          t[t[c].p[i]].B = t[t[c].p[i]].d;
        else
          t[t[c].p[i]].B = t[temp].B;
        q[wq++] = t[c].p[i];
      }
    }
    rq++;
  }
}
void solve() {
  memset(dp, 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < w; j++) {
      for (int k = 0; k < 10; k++) {
        if (dp[i][j][k]) {
          for (int l = 0; l < 4; l++) {
            int temp = j;
            while (temp != -1) {
              if (t[temp].p[l]) {
                temp = t[temp].p[l];
                break;
              }
              temp = t[temp].fail;
            }
            if (temp == -1) temp = 0;
            if (t[temp].d >= k + 1) {
              if (t[temp].B >= k + 1) {
                dp[i + 1][temp][0] += dp[i][j][k];
                dp[i + 1][temp][0] %= 1000000009;
              } else {
                dp[i + 1][temp][k + 1] += dp[i][j][k];
                dp[i + 1][temp][k + 1] %= 1000000009;
              }
            }
          }
        }
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  w = 0;
  build();
  for (int i = 0; i < m; i++) {
    scanf("%s", s[i]);
    l[i] = strlen(s[i]);
    ins(i);
  }
  ac();
  solve();
  out = 0;
  for (int i = 0; i < w; i++) {
    out += dp[n][i][0];
    out %= 1000000009;
  }
  printf("%d\n", out);
  return 0;
}
