"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;
long long read() {
  long long f = 1, x = 0;
  char c = getchar();
  while (c < '0' || '9' < c) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return f * x;
}
map<char, int> Map;
int ncnt, Root, nxt[1000 + 5][26];
int fail[1000 + 5], dep[1000 + 5];
int Newnode() {
  ++ncnt, fail[ncnt] = 0, dep[ncnt] = 0;
  memset(nxt[ncnt], 0, sizeof(nxt[ncnt]));
  return ncnt;
}
void Init() {
  ncnt = 0, Root = Newnode();
  return;
}
bool vis[1000 + 5];
void Insert(char *s, int len) {
  int u = Root;
  for (int i = 1; i <= len; i++) {
    int c = Map[s[i]];
    if (!nxt[u][c]) nxt[u][c] = Newnode();
    u = nxt[u][c];
  }
  dep[u] = max(dep[u], len);
  return;
}
void Make_fail() {
  queue<int> Q;
  for (int c = 0; c < 4; c++)
    if (nxt[Root][c])
      fail[nxt[Root][c]] = Root, Q.push(nxt[Root][c]);
    else
      nxt[Root][c] = Root;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    dep[u] = max(dep[u], dep[fail[u]]);
    for (int c = 0; c < 4; c++)
      if (nxt[u][c])
        fail[nxt[u][c]] = nxt[fail[u]][c], Q.push(nxt[u][c]);
      else
        nxt[u][c] = nxt[fail[u]][c];
  }
  return;
}
int Add(int x, int y) {
  x += y;
  return x >= (int)(1e9 + 9) ? x - (int)(1e9 + 9) : x;
}
char str[1030 + 5];
int a[1030 + 5];
int f[1030 + 5][1000 + 5][15];
int main() {
  Init();
  Map['A'] = 0, Map['G'] = 1, Map['C'] = 2, Map['T'] = 3;
  int n = read(), m = read();
  for (int i = 1; i <= m; i++)
    scanf("%s", str + 1), Insert(str, strlen(str + 1));
  Make_fail();
  f[0][Root][0] = 1;
  for (int i = 0; i < n; i++)
    for (int u = 1; u <= ncnt; u++)
      for (int j = 0; j <= 10; j++)
        for (int c = 0; c < 4; c++) {
          if (j + 1 <= dep[nxt[u][c]])
            f[i + 1][nxt[u][c]][0] = Add(f[i + 1][nxt[u][c]][0], f[i][u][j]);
          else
            f[i + 1][nxt[u][c]][j + 1] =
                Add(f[i + 1][nxt[u][c]][j + 1], f[i][u][j]);
        }
  int ans = 0;
  for (int u = 1; u <= ncnt; u++) ans = Add(ans, f[n][u][0]);
  printf("%d\n", ans);
  return 0;
}
