"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;
inline long long read() {
  long long f = 1, sum = 0;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return sum * f;
}
const int Mod = 1e9 + 9;
const int MAXN = 1010;
const int MAXM = 15;
const int CHARSET = 4;
map<char, int> mp;
char ss[MAXM][MAXN];
int tr[MAXM][CHARSET], fail[MAXN], size, len[MAXM];
int f[MAXN][MAXM * MAXM][MAXM];
int pre[MAXN], tag[MAXN];
void init() {
  size = 1;
  memset(tr[size], 0, sizeof(tr[size]));
  fail[size] = 0;
}
void new_node(int x, int ch) {
  tr[x][ch] = ++size;
  memset(tr[size], 0, sizeof(tr[size]));
  fail[size] = 0;
}
void add(int root, char s[], int num) {
  len[num] = strlen(s);
  int now = root;
  for (int i = 0; i < len[num]; i++) {
    int x = mp[s[i]];
    if (!tr[now][x]) new_node(now, x);
    now = tr[now][x];
  }
  tag[now] = num;
}
queue<int> q;
void get_fail(int root) {
  while (!q.empty()) q.pop();
  for (int i = 0; i < CHARSET; i++) {
    if (tr[root][i])
      fail[tr[root][i]] = root, q.push(tr[root][i]),
      pre[tr[root][i]] = tag[tr[root][i]];
    else
      tr[root][i] = root;
  }
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < CHARSET; i++) {
      if (tr[x][i]) {
        fail[tr[x][i]] = tr[fail[x]][i];
        q.push(tr[x][i]);
        if (tag[tr[x][i]])
          pre[tr[x][i]] = tag[tr[x][i]];
        else
          pre[tr[x][i]] = pre[tr[fail[x]][i]];
      } else
        tr[x][i] = tr[fail[x]][i];
    }
  }
}
int main() {
  mp['A'] = 0, mp['T'] = 1, mp['C'] = 2, mp['G'] = 3;
  int n, pp;
  scanf("%d%d", &n, &pp);
  init();
  for (int i = 1; i <= pp; i++) scanf("%s", ss[i]), add(1, ss[i], i);
  get_fail(1);
  f[0][1][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 1; j <= size; j++)
      for (int k = 0; k < 10; k++) {
        if (!f[i][j][k]) continue;
        for (int x = 0; x < CHARSET; x++) {
          int y = tr[j][x];
          if (len[pre[y]] >= k + 1)
            f[i + 1][y][0] = (f[i + 1][y][0] + f[i][j][k]) % Mod;
          else
            f[i + 1][y][k + 1] = (f[i + 1][y][k + 1] + f[i][j][k]) % Mod;
        }
      }
  int ans = 0;
  for (int i = 1; i <= size; i++) ans = (ans + f[n][i][0]) % Mod;
  cout << ans;
  return 0;
}
