"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 cc = 1024;
const int mod = 1000000009;
const int c = 16;
int n, m;
vector<string> vs;
set<string> st;
map<string, int> sw;
int p[c * c][4];
int a[cc][c][c * c];
int b[c * c];
inline char ch(int a) {
  if (a == 0) return 'A';
  if (a == 1) return 'C';
  if (a == 2) return 'G';
  if (a == 3) return 'T';
  return 0;
}
int main() {
  char tmp[c];
  char h;
  int i;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= m; ++i) {
    do h = getchar();
    while (h != '\n');
    scanf("%s", tmp);
    st.insert(string(tmp));
  }
  set<string>::iterator it;
  for (it = st.begin(); it != st.end(); ++it) {
    for (i = 0; i <= (*it).size(); ++i) vs.push_back((*it).substr(0, i));
  }
  sort(vs.begin(), vs.end());
  vs.resize(unique(vs.begin(), vs.end()) - vs.begin());
  string tt, stt;
  int j, t, w;
  for (i = 0; i < vs.size(); ++i)
    for (it = st.begin(); it != st.end(); ++it)
      if (vs[i].size() >= (*it).size() &&
          vs[i].substr(vs[i].size() - (*it).size(), (*it).size()) == *it)
        if (b[i] < (*it).size()) b[i] = (*it).size();
  for (i = 0; i < vs.size(); ++i) sw[vs[i]] = i;
  for (i = 0; i < vs.size(); ++i) {
    for (j = 0; j < 4; ++j) {
      tt = vs[i] + ch(j);
      for (t = tt.size(); t >= 1; --t) {
        stt = tt.substr(tt.size() - t, t);
        if (sw.count(stt)) {
          p[i][j] = sw[stt];
          break;
        }
      }
      if (t < 1) p[i][j] = -1;
    }
  }
  a[0][0][0] = 1;
  int nw;
  for (i = 0; i < n; ++i)
    for (w = 0; w < c; ++w)
      for (j = 0; j < vs.size(); ++j)
        for (t = 0; t < 4; ++t)
          if (p[j][t] != -1 && vs[p[j][t]].size() >= w + 1) {
            if (b[p[j][t]] >= w + 1)
              nw = 0;
            else
              nw = w + 1;
            if (nw < c)
              a[i + 1][nw][p[j][t]] =
                  (a[i + 1][nw][p[j][t]] + a[i][w][j]) % mod;
          }
  int ans = 0;
  for (i = 0; i < vs.size(); ++i) ans = (ans + a[n][0][i]) % mod;
  printf("%d\n", ans);
  return 0;
}
