"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;
struct Node {
  Node *go[26];
  bool leaf;
  Node *link;
  Node *anc;
  char to_anc;
  Node *clink;
  int depth;
  Node() {
    leaf = false;
    link = nullptr;
    anc = nullptr;
    clink = nullptr;
    depth = 0;
  }
} t[1010];
Node *root = t;
int tree_size = 1;
Node *go(Node *v, char c);
Node *link(Node *v);
Node *clink(Node *v);
Node *go(Node *v, char c) {
  if (v->go[c] == nullptr) {
    if (v == root) return v->go[c] = root;
    return v->go[c] = go(link(v), c);
  }
  return v->go[c];
}
Node *link(Node *v) {
  if (v->link == nullptr) {
    if (v == root || v->anc == root) return v->link = root;
    return v->link = go(link(v->anc), v->to_anc);
  }
  return v->link;
}
Node *clink(Node *v) {
  if (v->clink == nullptr) {
    if (v == root) return v->clink = root;
    if (link(v)->leaf) return v->clink = v->link;
    return v->clink = clink(v->link);
  }
  return v->clink;
}
void add(Node *v, Node *to, char c) {
  v->go[c] = to;
  to->anc = v;
  to->to_anc = c;
  to->depth = v->depth + 1;
}
int maxD(Node *v) {
  if (v->leaf) return v->depth;
  if (v == root) return 0;
  return clink(v)->depth;
}
void add(const string &s) {
  Node *v = root;
  for (char ch : s) {
    int c = ch - 'A';
    if (v->go[c] == nullptr) {
      add(v, t + tree_size, c);
      ++tree_size;
    }
    v = v->go[c];
  }
  v->leaf = true;
}
int n, m;
int d[1010][13], h[1010][13];
void read() {
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    string s;
    cin >> s;
    add(s);
  }
}
void next(int x, int y, char c, int &tx, int &ty) {
  tx = (int)(go(t + x, c) - t);
  ty = y + 1;
  int d = maxD(t + tx);
  if (d >= ty) ty = 0;
}
void din_step() {
  for (int i = 0; i < tree_size; ++i)
    for (int j = 0; j < 13; ++j)
      if (d[i][j]) {
        int tx, ty;
        for (int k = 0; k < 26; ++k) {
          next(i, j, k, tx, ty);
          if (ty >= 13) continue;
          h[tx][ty] += d[i][j];
          if (h[tx][ty] >= 1000000009) h[tx][ty] -= 1000000009;
        }
      }
  for (int i = 0; i < tree_size; ++i)
    for (int j = 0; j < 13; ++j) {
      d[i][j] = h[i][j];
      h[i][j] = 0;
    }
}
int main() {
  read();
  d[0][0] = 1;
  for (int i = 0; i < n; ++i) din_step();
  int ans = 0;
  for (int i = 0; i < tree_size; ++i) {
    ans += d[i][0];
    if (ans >= 1000000009) ans -= 1000000009;
  }
  cout << ans << "\n";
  return 0;
}
