"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;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while ((c < '0' || c > '9') && c - '-') c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - 48;
    c = getchar();
  }
  return x * f;
}
const int N = 1e3 + 10;
struct pig {
  int fail;
  int c[5];
} tre[N];
int pos = 0, sl = 0, n, dp[N][N][11];
int m, val[N];
char c[N];
const int mo = 1e9 + 9;
inline int dui(char ch) {
  if (ch == 'A') return 4;
  if (ch == 'C') return 3;
  if (ch == 'G') return 2;
  if (ch == 'T') return 1;
}
inline int max(int x, int y) { return x < y ? y : x; }
inline int min(int x, int y) { return x < y ? x : y; }
int duo;
inline void cl() {
  int len = strlen(c + 1);
  duo = max(duo, len), pos = 0;
  for (int i = 1; i <= len; i++) {
    int g = dui(c[i]);
    if (!tre[pos].c[g]) tre[pos].c[g] = ++sl;
    pos = tre[pos].c[g];
  }
  val[pos] = max(val[pos], len);
}
inline void get_fail() {
  queue<int> que;
  for (int i = 1; i <= 4; i++)
    if (tre[0].c[i]) que.push(tre[0].c[i]);
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    for (int i = 1; i <= 4; i++) {
      int v = tre[u].c[i];
      if (v)
        tre[v].fail = tre[tre[u].fail].c[i], que.push(v),
        val[v] = max(val[v], val[tre[v].fail]);
      else
        tre[u].c[i] = tre[tre[u].fail].c[i];
    }
  }
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= m; i++) scanf("%s", c + 1), cl();
  get_fail();
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= sl; j++)
      for (int k = 0; k <= duo; k++) {
        int lun = dp[i][j][k];
        if (lun) {
          for (int er = 1; er <= 4; er++) {
            int xin = tre[j].c[er];
            (dp[i + 1][xin][val[xin] > k ? 0 : min(k + 1, duo)] +=
             dp[i][j][k]) %= mo;
          }
        }
      }
  int dan = 0;
  for (int i = 1; i <= sl; i++) dan = (dan + dp[n][i][0]) % mo;
  cout << dan << "\n";
  return 0;
}
