"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;
struct _ {
  _() { ios_base::sync_with_stdio(0); }
} _;
template <class A, class B>
ostream &operator<<(ostream &o, pair<A, B> t) {
  o << "(" << t.first << ", " << t.second << ")";
  return o;
}
template <class T>
string tostring(T first, int len = 0, char c = '0') {
  stringstream ss;
  ss << first;
  string r = ss.str();
  if (r.length() < len) r = string(len - r.length(), c) + r;
  return r;
}
template <class T>
void PV(T a, T b, int n = 0, int w = 0, string s = " ") {
  int c = 0;
  while (a != b) {
    cout << tostring(*a++, w, ' ');
    if (a != b && (n == 0 || ++c % n))
      cout << s;
    else
      cout << "\n";
    cout.flush();
  }
}
template <class T>
inline bool chmin(T &a, T b) {
  return a > b ? a = b, 1 : 0;
}
template <class T>
inline bool chmax(T &a, T b) {
  return a < b ? a = b, 1 : 0;
}
const long long linf = 0x3f3f3f3f3f3f3f3fLL;
const int inf = 0x3f3f3f3f;
const int mod = int(1e9) + 9;
const int N = 1111;
int n, m;
string s[11];
map<string, int> prefixForID;
string prefixByID[111];
int go[111][4];
map<char, char> M;
int dp[N][11][111];
int finish[111];
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
int main() {
  M['A'] = 'a';
  M['C'] = 'b';
  M['G'] = 'c';
  M['T'] = 'd';
  cin >> n >> m;
  int ID = 0;
  prefixForID[""] = ID++;
  for (int i = 0; i < m; i++) {
    cin >> s[i];
    for (char &c : s[i]) c = M[c];
    string p;
    for (int j = 0; j < s[i].length(); j++) {
      p.push_back(s[i][j]);
      if (prefixForID.find(p) == prefixForID.end()) {
        prefixForID[p] = ID;
        prefixByID[ID] = p;
        ID++;
      }
    }
  }
  for (int i = 0; i < ID; i++) {
    string p = prefixByID[i];
    for (int j = 0; j < m; j++) {
      if (s[j].length() <= p.length()) {
        if (s[j] == p.substr(p.length() - s[j].length()))
          chmax(finish[i], (int)s[j].length());
      }
    }
  }
  for (int i = 0; i < ID; i++)
    for (int j = 0; j < 4; j++) {
      string p = prefixByID[i] + char('a' + j);
      while (prefixForID.find(p) == prefixForID.end()) p = p.substr(1);
      if (i == 0 && j == 3)
        cerr << "["
                "prefixForID[p]"
             << " = " << (prefixForID[p]) << "]" << endl;
      ;
      go[i][j] = prefixForID[p];
    }
  dp[0][0][0] = 1;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 10; j++)
      for (int k = 0; k < ID; k++) {
        int first = dp[i][j][k];
        if (first == 0) continue;
        for (int w = 0; w < 4; w++) {
          int to = go[k][w];
          if (finish[to] >= j + 1) {
            add(dp[i + 1][0][to], first);
          } else {
            if (j + 1 < 10) add(dp[i + 1][j + 1][to], first);
          }
        }
      }
  cerr << "["
          "go[0][3]"
       << " = " << (go[0][3]) << "]" << endl;
  ;
  cerr << "["
          "dp[1][0][prefixForID[\"d\"]]"
       << " = " << (dp[1][0][prefixForID["d"]]) << "]" << endl;
  ;
  int res = 0;
  for (int k = 0; k < ID; k++) add(res, dp[n][0][k]);
  cout << res << endl;
  return 0;
}
