"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 mmod = 1e9 + 9;
int N, M, len, dp[1001][101][11], ans;
char ss[20];
struct Node {
  int maxLen, id;
  int cc;
  Node *ch[4], *fail;
  Node() {
    maxLen = 0;
    ch[0] = ch[1] = ch[2] = ch[3] = fail = NULL;
  }
} w[500], *root, *tw = w;
void Insert() {
  Node *nd = root;
  for (int i = 0; i < len; i++) {
    if (!nd->ch[ss[i]]) {
      nd->ch[ss[i]] = ++tw;
      nd->ch[ss[i]]->id = tw - w;
      nd->ch[ss[i]]->cc = ss[i];
    }
    nd = nd->ch[ss[i]];
  }
  nd->maxLen = len;
}
Node *q[500];
int ba, fr, vis[505];
void getFail() {
  ba = 0;
  fr = 1;
  root->fail = root;
  q[++ba] = root;
  vis[root->id] = true;
  while (ba >= fr) {
    Node *u = q[fr++];
    for (int i = 0; i < 4; i++) {
      Node *v = u->ch[i];
      if (!v || v->fail) continue;
      if (u == root)
        v->fail = root;
      else {
        Node *p = u->fail;
        while (!p->ch[i] && p != root) p = p->fail;
        v->fail = (p->ch[i] ? p->ch[i] : root);
        v->maxLen = max(v->maxLen, v->fail->maxLen);
      }
      for (int j = 0; j < 4; j++)
        if (!v->ch[j]) v->ch[j] = v->fail->ch[j];
      q[++ba] = v;
    }
  }
}
void solve() {
  dp[0][1][0] = 1;
  for (int i = 0; i < N; i++) {
    for (int j = 1; j <= tw - w; j++) {
      Node *nd = &w[j];
      for (int k = 0; k <= 10; k++) {
        if (!dp[i][j][k]) continue;
        for (int sn = 0; sn < 4; sn++) {
          Node *nxt = nd->ch[sn];
          if (!nxt) continue;
          if (nxt->maxLen >= k + 1) {
            dp[i + 1][nxt->id][0] =
                (dp[i + 1][nxt->id][0] + dp[i][j][k]) % mmod;
          } else if (k != 10) {
            dp[i + 1][nxt->id][k + 1] =
                (dp[i + 1][nxt->id][k + 1] + dp[i][j][k]) % mmod;
          }
        }
      }
    }
  }
  for (int j = 1; j <= tw - w; j++) ans = (ans + dp[N][j][0]) % mmod;
  printf("%d", ans);
}
int main() {
  scanf("%d%d", &N, &M);
  root = ++tw;
  root->id = tw - w;
  for (int i = 1; i <= M; i++) {
    scanf("%s", ss);
    len = strlen(ss);
    for (int j = 0; j < len; j++)
      if (ss[j] == 'A')
        ss[j] = 0;
      else if (ss[j] == 'T')
        ss[j] = 1;
      else if (ss[j] == 'C')
        ss[j] = 2;
      else if (ss[j] == 'G')
        ss[j] = 3;
    Insert();
  }
  getFail();
  solve();
}
