"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 = 110, M = 1010, mod = 1e9 + 9;
int n, m, tot, ans;
int tag[N], dep[N], fail[N], lst[N], ch[N][4], dp[M][N][15];
char s[15];
queue<int> q;
inline void addm(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int idx(char c) {
  if (c == 'A') return 0;
  if (c == 'T') return 1;
  if (c == 'C') return 2;
  return 3;
}
void ins(char *s) {
  int len = strlen(s + 1), now = 0;
  for (int i = 1; i <= len; i++) {
    int nxt = idx(s[i]);
    if (!ch[now][nxt]) ch[now][nxt] = ++tot, dep[tot] = dep[now] + 1;
    now = ch[now][nxt];
  }
  tag[now] = 1;
}
void build() {
  for (int i = 0; i < 4; i++)
    if (ch[0][i]) fail[ch[0][i]] = 0, lst[ch[0][i]] = 0, q.push(ch[0][i]);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int i = 0; i < 4; i++) {
      if (ch[u][i]) {
        fail[ch[u][i]] = ch[fail[u]][i];
        q.push(ch[u][i]);
      } else
        ch[u][i] = ch[fail[u]][i];
    }
    lst[u] = (tag[u]) ? u : lst[fail[u]];
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%s", s + 1), ins(s);
  build();
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= tot; j++)
      for (int k = 0; k <= 10; k++) {
        if (!dp[i][j][k]) continue;
        for (int nxt = 0; nxt < 4; nxt++) {
          int v = ch[j][nxt], tp = k + 1;
          if (dep[lst[v]] >= k + 1) tp = 0;
          addm(dp[i + 1][v][tp], dp[i][j][k]);
        }
      }
  for (int i = 0; i <= tot; i++) addm(ans, dp[n][i][0]);
  printf("%d", ans);
}
