"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 = 1e2 + 5;
const int mod = 1e9 + 9;
int getc(char c) {
  if (c == 'A') return 0;
  if (c == 'G') return 1;
  if (c == 'T') return 2;
  return 3;
}
struct node {
  int next[4], fail, val;
} st[maxn];
int sz = 0;
vector<int> g[maxn];
void insert(char *s) {
  int now = 0;
  for (int i = 1; s[i]; i++) {
    if (!st[now].next[getc(s[i])]) st[now].next[getc(s[i])] = ++sz;
    now = st[now].next[getc(s[i])];
  }
  st[now].val = strlen(s + 1);
}
void dfs(int x) {
  st[x].val = max(st[x].val, st[st[x].fail].val);
  for (int i = (0); i < (g[x].size()); i++) dfs(g[x][i]);
}
void build() {
  queue<int> q;
  for (int i = 0; i < 4; i++)
    if (st[0].next[i]) q.push(st[0].next[i]);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      if (st[x].next[i]) {
        st[st[x].next[i]].fail = st[st[x].fail].next[i];
        q.push(st[x].next[i]);
      } else
        st[x].next[i] = st[st[x].fail].next[i];
    }
  }
  for (int i = (1); i <= (sz); i++) g[st[i].fail].push_back(i);
  dfs(0);
}
char s[maxn];
int f[maxn][maxn * 10][12], n;
int dfs(int now, int len, int k) {
  if (k > 10) return 0;
  if (f[now][len][k] != -1) return f[now][len][k];
  if (len == n) return (k == 0);
  int ans = 0;
  for (int i = (0); i <= (3); i++) {
    int v = st[now].next[i];
    if (st[v].val >= k + 1)
      (ans += dfs(v, len + 1, 0)) %= mod;
    else
      (ans += dfs(v, len + 1, k + 1)) %= mod;
  }
  return f[now][len][k] = ans;
}
int main() {
  memset(f, -1, sizeof(f));
  int m;
  scanf("%d", &n);
  scanf("%d", &m);
  while (m--) {
    scanf("%s", s + 1);
    insert(s);
  }
  build();
  cout << dfs(0, 0, 0);
}
