"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 inf = 1000 * 1000 * 1000;
vector<string> pref;
vector<string> st;
int n, m;
int dp[1 << 10][11][1 << 7];
int fp[1 << 7][4];
const int MOD = inf + 9;
const string s = "ACGT";
int findp(int p, int c) {
  if (fp[p][c] == inf) {
    fp[p][c] = -1;
    int res = -1;
    string ss = pref[p] + s[c];
    for (int i = 0; i < pref.size(); ++i)
      if (pref[i].size() <= ss.size())
        if (ss.substr(ss.size() - pref[i].size(), pref[i].size()) == pref[i])
          if (res < (int)pref[i].size()) res = pref[i].size(), fp[p][c] = i;
    if (res == 0) fp[p][c] = -1;
  }
  return fp[p][c];
}
int ff(int k, int p) {
  int res = k;
  for (int i = 0; i < (int)pref[p].size() - k + 1; ++i)
    for (int j = 0; j < st.size(); ++j)
      if (st[j].size() + i <= pref[p].size())
        if (pref[p].substr(i, st[j].size()) == st[j])
          res = min(res, (int)pref[p].size() - i - (int)st[j].size());
  return res;
}
int f(int pos, int k, int p) {
  if (pos > n) return 0;
  if (pos == n) return k == 0;
  if (k > pref[p].size()) return 0;
  if (dp[pos][k][p] == -1) {
    int& res = dp[pos][k][p];
    res = 0;
    for (int i = 0; i < 4; ++i) {
      int newp = findp(p, i);
      if (newp == -1) continue;
      int rem = ff(k + 1, newp);
      res = (res + f(pos + 1, rem, newp)) % MOD;
    }
  }
  return dp[pos][k][p];
}
int main() {
  for (int i = 0; i < (1 << 7); ++i)
    for (int j = 0; j < 4; ++j) fp[i][j] = inf;
  memset(dp, -1, sizeof(dp));
  cin >> n >> m;
  pref.push_back("");
  for (int i = 0; i < m; ++i) {
    string s;
    cin >> s;
    st.push_back(s);
    string temp = "";
    for (int j = 0; j < s.size(); ++j) temp += s[j], pref.push_back(temp);
  }
  cout << f(0, 0, 0);
  return 0;
}
