"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 mod = 1e9 + 9;
inline int num(char c) {
  if (c == 'A') return 0;
  if (c == 'G') return 1;
  if (c == 'C') return 2;
  return 3;
}
const int N = 1010;
const int M = 110;
int n, m;
char s[20];
struct Aho_Corasick_Automaton {
  long long ch[M][4], fail[M], val[M], dp[N][M][15], ans, tail;
  queue<int> Q;
  inline void clear() {
    memset(ch, 0, sizeof(ch));
    memset(fail, 0, sizeof(fail));
    memset(val, 0, sizeof(val));
    memset(dp, 0, sizeof(dp));
    while (!Q.empty()) Q.pop();
    tail = ans = 0;
  }
  inline void insert() {
    int len = strlen(s);
    int tmp = 0;
    for (int i = 0; i < len; i++) {
      int c = num(s[i]);
      if (!ch[tmp][c]) ch[tmp][c] = ++tail;
      tmp = ch[tmp][c];
    }
    val[tmp] = len;
  }
  inline void getfail() {
    for (int i = 0; i < 4; i++)
      if (ch[0][i]) {
        Q.push(ch[0][i]);
        fail[ch[0][i]] = 0;
      }
    while (!Q.empty()) {
      int top = Q.front();
      Q.pop();
      for (int i = 0; i < 4; i++) {
        if (!ch[top][i]) {
          ch[top][i] = ch[fail[top]][i];
          continue;
        }
        int u = ch[top][i];
        fail[u] = ch[fail[top]][i];
        val[u] = max(val[u], val[fail[u]]);
        Q.push(u);
      }
    }
  }
  inline void getans() {
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    for (int i = 0; i < n; i++)
      for (int u = 0; u <= tail; u++)
        for (int j = 0; j <= 10; j++)
          for (int c = 0; c < 4; c++) {
            int v = ch[u][c];
            if (val[v] >= j + 1)
              dp[i + 1][v][0] = (dp[i + 1][v][0] + dp[i][u][j]) % mod;
            else
              dp[i + 1][v][j + 1] = (dp[i][u][j] + dp[i + 1][v][j + 1]) % mod;
          }
    long long ans = 0;
    for (int i = 1; i <= tail; i++) ans = (ans + dp[n][i][0]) % mod;
    printf("%lld\n", ans);
  }
} AC;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", s);
    AC.insert();
  }
  AC.getfail();
  AC.getans();
  return 0;
}
