"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 MAXN = 1005;
const int LEN = 4;
const long long MOD = 1e9 + 9;
struct node {
  int id;
  int layer;
  int cnt;
  node* nxt[LEN];
  bool have[26];
  node* fail;
  int low;
  node() : id(0), cnt(0), fail(NULL), low(0), layer(0) {
    memset(nxt, 0, sizeof(nxt));
    memset(have, 0, sizeof(have));
  }
};
node nodes[MAXN];
int tot = 1;
node* root = nodes;
node* newNode() {
  nodes[tot].id = tot;
  return &nodes[tot++];
}
char mp[256];
void insertS(char* str) {
  node* nd = root;
  for (int i = 0; str[i]; i++) {
    int id = mp[str[i]];
    if (!nd->nxt[id]) {
      nd->nxt[id] = newNode();
      nd->nxt[id]->layer = i + 1;
    }
    nd = nd->nxt[id];
  }
  nd->cnt = 1;
}
void build() {
  queue<node*> que;
  root->fail = root;
  for (int i = 0; i < LEN; i++) {
    if (!root->nxt[i]) {
      root->nxt[i] = root;
    } else {
      root->nxt[i]->fail = root;
      root->have[i] = true;
      que.push(root->nxt[i]);
    }
  }
  while (!que.empty()) {
    node* nd = que.front();
    que.pop();
    nd->low = nd->fail->low;
    if (nd->cnt) {
      nd->low = nd->layer;
    }
    for (int i = 0; i < LEN; i++) {
      if (nd->nxt[i]) {
        nd->nxt[i]->fail = nd->fail->nxt[i];
        nd->have[i] = true;
        que.push(nd->nxt[i]);
      } else {
        nd->nxt[i] = nd->fail->nxt[i];
      }
    }
  }
}
int n, m;
char str[MAXN];
long long dp[MAXN][105][11];
int main() {
  mp['A'] = 0;
  mp['C'] = 1;
  mp['G'] = 2;
  mp['T'] = 3;
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    scanf("%s", str);
    insertS(str);
  }
  build();
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < tot; j++) {
      for (int l = 0; l <= 10; l++) {
        for (int k = 0; k < LEN; k++) {
          if (dp[i][j][l]) {
            if (nodes[j].layer - nodes[j].low <= l) {
              if (nodes[j].nxt[k] != root) {
                long long& loc =
                    dp[i + 1][nodes[j].nxt[k]->id][nodes[j].nxt[k]->layer - 1];
                loc = (loc + dp[i][j][l]) % MOD;
              }
            } else {
              if (nodes[j].nxt[k]->layer + l > nodes[j].layer &&
                  nodes[j].nxt[k] != root) {
                long long& loc =
                    dp[i + 1][nodes[j].nxt[k]->id]
                      [nodes[j].nxt[k]->layer + l - nodes[j].layer - 1];
                loc = (loc + dp[i][j][l]) % MOD;
              }
            }
          }
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 0; i < tot; i++) {
    for (int j = 0; j <= 10; j++) {
      if (nodes[i].layer - nodes[i].low <= j) {
        ans = (ans + dp[n][i][j]) % MOD;
      }
    }
  }
  printf("%I64d\n", ans);
  return 0;
}
