"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;
string ss, s;
const int md = ((int)1e9) + 9;
int n, kv, m, ans;
int b[1000][250];
int sf[1000];
int pr[1000];
int fl[1000];
int len[1000];
int lenn[1000];
int dp[2000][20][200];
char prc[1000];
void add(string s) {
  int tk = 1;
  for (int i = 0; i < (int)s.length(); i++) {
    if (b[tk][s[i]] == 0) {
      b[tk][s[i]] = ++kv;
      pr[kv] = tk;
      len[kv] = len[tk] + 1;
      prc[kv] = s[i];
    }
    tk = b[tk][s[i]];
  }
  fl[tk] = 1;
  lenn[tk] = len[tk];
}
void calcsf(int v) {
  if (sf[v] != -1) return;
  if (pr[v] == 1)
    sf[v] = 1;
  else {
    calcsf(pr[v]);
    int tk = sf[pr[v]];
    while (tk != 1 && b[tk][prc[v]] == 0) {
      calcsf(tk);
      tk = sf[tk];
    }
    if (b[tk][prc[v]] != 0)
      sf[v] = b[tk][prc[v]];
    else
      sf[v] = 1;
  }
  calcsf(sf[v]);
  if (fl[sf[v]] == 1) {
    fl[v] = 1;
    lenn[v] = max(lenn[v], lenn[sf[v]]);
  }
}
int main() {
  scanf("%d%d", &n, &m);
  kv = 1;
  for (int i = 0; i < m; i++) {
    cin >> s;
    add(s);
  }
  for (int i = 1; i <= kv; i++) {
    sf[i] = -1;
  }
  sf[1] = 1;
  len[1] = 0;
  for (int i = 2; i <= kv; i++) calcsf(i);
  ss = "ACGT";
  for (int i = 2; i <= kv; i++)
    for (int j = 0; j < 4; j++) {
      int tk = i;
      while (tk != 1 && b[tk][ss[j]] == 0) {
        tk = sf[tk];
      }
      if (b[tk][ss[j]] != 0) {
        b[i][ss[j]] = b[tk][ss[j]];
      }
    }
  dp[0][0][1] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j <= 10; j++) {
      for (int k = 1; k <= kv; k++)
        if (dp[i][j][k] != 0) {
          for (int l = 0; l < 4; l++)
            if (b[k][ss[l]] != 0) {
              int hh = j + 1;
              if (fl[b[k][ss[l]]] == 1 && lenn[b[k][ss[l]]] >= hh) {
                hh = 0;
              }
              dp[i + 1][hh][b[k][ss[l]]] =
                  (dp[i + 1][hh][b[k][ss[l]]] + dp[i][j][k]) % md;
            }
        }
    }
  ans = 0;
  for (int i = 1; i <= kv; i++)
    if (fl[i] == 1) {
      ans = (ans + dp[n][0][i]) % md;
    }
  cout << ans << endl;
  return 0;
}
