"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 P = 1e9 + 9;
int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a < 0 ? a + P : a;
}
int mul(int a, int b) { return 1ll * a * b % P; }
int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 1e5 + 6;
int a[N], len, n, m, f[1005][305][20], ans;
char s[N];
int get(int x) {
  if (x == 'A') return 0;
  if (x == 'G') return 1;
  if (x == 'C') return 2;
  if (x == 'T') return 3;
}
struct Trie {
  static const int N = 101010, M = 4;
  int ne[N][M], fail[N], fa[N], rt, L;
  void ini() {
    fill_n(ne[fail[0] = N - 1], M, 0);
    L = 0;
    rt = newnode();
  }
  int newnode() {
    fill_n(ne[L], M, 0);
    return L++;
  }
  void add(char *s) {
    int p = rt;
    for (int i = 0; s[i]; ++i) {
      int c = get(s[i]);
      if (!ne[p][c]) ne[p][c] = newnode(), fa[L - 1] = p;
      p = ne[p][c];
    }
    a[p] = max(a[p], len);
  }
  void Build() {
    vector<int> v;
    v.push_back(rt);
    for (int i = (0); i < ((int)v.size()); i++) {
      int c = v[i];
      for (int i = (0); i < (M); i++)
        ne[c][i]                            ? v.push_back(ne[c][i]),
            fail[ne[c][i]] = ne[fail[c]][i] : ne[c][i] = ne[fail[c]][i];
    }
  }
} tr;
int main() {
  cin >> n >> m;
  tr.ini();
  for (int i = (1); i < (m + 1); i++) {
    cin >> s;
    len = strlen(s);
    tr.add(s);
  }
  tr.Build();
  f[0][0][0] = 1;
  for (int i = (1); i < (n + 1); i++) {
    for (int j = (0); j < (tr.L); j++)
      for (int l = (0); l < (4); l++) {
        int v = tr.ne[j][l], u = v, t = 0;
        while (u) {
          t = max(t, a[u]);
          u = tr.fail[u];
        }
        for (int k = (0); k < (t); k++)
          f[i][v][0] = add(f[i][v][0], f[i - 1][j][k]);
        for (int k = (t); k < (11); k++)
          f[i][v][k + 1] = add(f[i][v][k + 1], f[i - 1][j][k]);
      }
  }
  for (int j = (0); j < (tr.L); j++) ans = add(ans, f[n][j][0]);
  cout << ans;
  return 0;
}
