"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;
int dp[1005][105][15];
char s[1005];
int son[105][26], cnt, f[105], val[105], dep[105];
inline void insert(char *s) {
  int len = strlen(s), t = 0;
  for (int i = 0; i < len; ++i) {
    int c = s[i] - 'A';
    if (!son[t][c]) {
      son[t][c] = ++cnt;
      dep[cnt] = dep[t] + 1;
    }
    t = son[t][c];
  }
  val[t] = dep[t];
}
void AC() {
  f[0] = 0;
  queue<int> q;
  for (int i = 0; i < 26; ++i)
    if (son[0][i]) f[son[0][i]] = 0, q.push(son[0][i]);
  while (!q.empty()) {
    int r = q.front();
    q.pop();
    for (int c = 0; c < 26; ++c) {
      int u = son[r][c];
      if (!u) {
        son[r][c] = son[f[r]][c];
        continue;
      }
      q.push(u);
      int v = f[r];
      f[u] = son[v][c];
      val[u] = max(val[u], val[f[u]]);
    }
  }
}
int solve(int n, int pos, int pre) {
  if (dp[n][pos][pre] != -1) return dp[n][pos][pre];
  if (n == 0) return dp[n][pos][pre] = (val[pos] >= pre);
  int ret = 0;
  for (int i = 0; i < 26; ++i) {
    int npos = son[pos][i], npre = val[pos] >= pre ? 1 : pre + 1;
    if (npos && dep[npos] >= npre)
      ret = (ret + solve(n - 1, npos, npre)) % 1000000009;
  }
  return dp[n][pos][pre] = ret;
}
int main() {
  int n, k;
  while (scanf("%d%d", &n, &k) == 2) {
    memset(son, 0, sizeof(son));
    memset(f, 0, sizeof(f));
    memset(val, 0, sizeof(val));
    memset(dep, 0, sizeof(dep));
    cnt = 0;
    for (int i = 0; i < k; ++i) {
      scanf("%s", s);
      insert(s);
    }
    AC();
    memset(dp, -1, sizeof(dp));
    cout << solve(n, 0, 0) << endl;
  }
  return 0;
}
