"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 = 333333;
int c[maxn * 10];
const int mod = 1000000009;
int dp[1111][111][15];
char s[15];
int l[15];
int get(char c) {
  if (c == 'A') return 0;
  if (c == 'C') return 1;
  if (c == 'G') return 2;
  return 3;
}
struct ac_auto {
  queue<int> Q;
  int tot, c[4][111];
  int val[111], fail[111];
  int new_node() {
    int i;
    for (i = 0; i < 4; i++) c[i][tot] = 0;
    fail[tot] = val[tot] = 0;
    return tot++;
  }
  void init() {
    tot = 0;
    new_node();
  }
  void insert(char *s, int id) {
    int now = 0;
    for (; *s; s++) {
      int k = get(*s);
      if (!c[k][now]) c[k][now] = new_node();
      now = c[k][now];
    }
    val[now] = l[id];
  }
  void get_fail() {
    int i, u = 0;
    for (i = 0; i < 4; i++)
      if (c[i][u]) Q.push(c[i][u]);
    while (!Q.empty()) {
      u = Q.front(), Q.pop();
      for (i = 0; i < 4; i++) {
        if (c[i][u]) {
          int e = c[i][u];
          int j = fail[u];
          fail[e] = c[i][j];
          val[e] = max(val[e], val[fail[e]]);
          Q.push(e);
        } else
          c[i][u] = c[i][fail[u]];
      }
    }
  }
  void solve(int n) {
    int i, j, k, p;
    for (i = 0; i < n; i++) {
      for (j = 0; j < tot; j++)
        for (k = 0; k <= 10; k++) {
          if (dp[i][j][k]) {
            for (p = 0; p < 4; p++) {
              int t = c[p][j];
              if (val[t] >= k + 1)
                dp[i + 1][t][0] = (dp[i + 1][t][0] + dp[i][j][k]) % mod;
              else
                dp[i + 1][t][k + 1] = (dp[i + 1][t][k + 1] + dp[i][j][k]) % mod;
            }
          }
        }
    }
    int ans = 0;
    for (i = 0; i < tot; i++) ans = (ans + dp[n][i][0]) % mod;
    printf("%d\n", ans);
  }
} ac;
int main() {
  int n, m, i, j, k, p, r, t;
  while (scanf("%d%d", &n, &m) != EOF) {
    ac.init();
    j = 0;
    while (m--) {
      scanf("%s", s);
      l[++j] = strlen(s);
      ac.insert(s, j);
    }
    ac.get_fail();
    memset(dp, 0, sizeof(dp));
    dp[0][0][0] = 1;
    ac.solve(n);
  }
}
