"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;
#pragma comment(linker, "/STACK:1024000000,1024000000")
const int MAX_NODE = 100 + 5, MAX_BRANCH = 4;
const int MAX_N = 1000 + 5, mod = 1000000009, MAX_LEN = 10;
int n, m, dp[MAX_N][MAX_NODE][MAX_LEN];
struct Node {
  Node *fl, *go[MAX_BRANCH];
  int id, acc;
  void reset(int indice) {
    id = indice;
    acc = 0;
    memset(go, 0, sizeof(go));
  }
};
Node node[MAX_NODE], *root, *que[MAX_NODE], *node0;
int sz, branch;
Node *newNode() {
  node[sz].reset(sz);
  return &node[sz++];
}
int cti[256];
namespace AC {
Node *cur;
void init() {
  sz = 0;
  root = newNode();
}
void insert(char *str, int len) {
  cur = root;
  for (int i = 0; i < len; i++) {
    int a = cti[str[i]];
    if (!cur->go[a]) cur->go[a] = newNode();
    cur = cur->go[a];
  }
  cur->acc = len;
}
void finish() {
  int ll = 0, rr = 0;
  root->fl = root;
  for (int i = 0; i < branch; i++)
    if (root->go[i]) {
      root->go[i]->fl = root;
      que[rr++] = root->go[i];
    } else
      root->go[i] = root;
  while (ll < rr) {
    cur = que[ll++];
    for (int i = 0; i < branch; i++) {
      if (cur->go[i]) {
        cur->go[i]->fl = cur->fl->go[i];
        cur->go[i]->acc = max(cur->go[i]->acc, cur->go[i]->fl->acc);
        que[rr++] = cur->go[i];
      } else
        cur->go[i] = cur->fl->go[i];
    }
  }
}
};  // namespace AC
void update(int &a, int upd) {
  if ((a += upd) >= mod) a -= mod;
}
int isDirectSon(Node *x, int j) {
  if (x == root)
    return x->go[j] != x;
  else
    return x->go[j] != x->fl->go[j];
}
void work() {
  memset(dp, 0, sizeof(dp));
  dp[0][root->id][0] = 1;
  for (int l = 0; l < n; l++)
    for (int i = 0; i < sz; i++)
      for (int k = 0; k < 10; k++)
        if (dp[l][i][k]) {
          for (int c = 0; c < branch; c++) {
            int j = node[i].go[c]->id;
            if (node[j].acc >= k + 1)
              update(dp[l + 1][j][0], dp[l][i][k]);
            else if (k + 1 < 10)
              update(dp[l + 1][j][k + 1], dp[l][i][k]);
          }
        }
  int res = 0;
  for (int i = 0; i < sz; i++)
    if (node[i].acc) update(res, dp[n][i][0]);
  printf("%d\n", res);
}
char str[100];
int main() {
  cti['A'] = 0;
  cti['C'] = 1;
  cti['G'] = 2;
  cti['T'] = 3;
  branch = 4;
  scanf("%d %d", &n, &m);
  AC::init();
  for (int i = 0; i < m; i++) {
    scanf("%s", str);
    AC::insert(str, strlen(str));
  }
  AC::finish();
  work();
  return 0;
}
