"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;
int nextInt() {
  int x;
  scanf("%d", &x);
  return x;
}
long long nextLong() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
double nextDouble() {
  double x;
  scanf("%lf", &x);
  return x;
}
const int BUFSIZE = 1000000;
char buf[BUFSIZE + 1];
string nextString() {
  scanf("%s", buf);
  return buf;
}
const int MOD = 1000000009;
const int WORDLENGTH = 10;
const int DICTSIZE = 10;
const int TRIESIZE = WORDLENGTH * DICTSIZE + 1;
const int LENGTH = 1000;
vector<string> dict;
map<string, int> prefs;
int getPref(const string &s) {
  map<string, int>::iterator it = prefs.find(s);
  if (it == prefs.end()) {
    return prefs[s] = prefs.size();
  }
  return it->second;
}
map<int, string> nextPrefMemo[26];
const string &nextPref(string pref, int c) {
  int id = getPref(pref);
  if (nextPrefMemo[c].find(id) != nextPrefMemo[c].end()) {
    return nextPrefMemo[c][id];
  }
  pref += c + 'a';
  string res;
  for (int len = pref.size(); len >= 0; --len) {
    string sub = pref.substr(pref.size() - len);
    for (int i = 0; i < dict.size(); ++i) {
      if (dict[i].size() >= len && dict[i].substr(0, len) == sub) {
        return nextPrefMemo[c][id] = sub;
      }
    }
  }
}
int maxWordInPref(const string &s) {
  for (int len = s.size(); len > 0; --len) {
    string sub = s.substr(s.size() - len);
    for (int i = 0; i < dict.size(); ++i) {
      if (dict[i] == sub) {
        return len;
      }
    }
  }
  return 0;
}
int memo[TRIESIZE][WORDLENGTH + 1][LENGTH + 1];
int solve(const string &pref, int notCover, int length) {
  if (length == 0) {
    return notCover == 0;
  }
  int &res = memo[getPref(pref)][notCover][length];
  if (res != -1) {
    return res;
  }
  res = 0;
  for (int c = 0; c < 26; ++c) {
    const string &npref = nextPref(pref, c);
    if (npref.empty()) {
      continue;
    }
    int nnotCover = notCover + 1;
    if (maxWordInPref(npref) >= nnotCover) {
      nnotCover = 0;
    }
    if (nnotCover > npref.size()) {
      continue;
    }
    res = (res + solve(npref, nnotCover, length - 1)) % MOD;
  }
  return res;
}
int main() {
  int n = nextInt();
  int m = nextInt();
  dict = vector<string>(m);
  for (int i = 0; i < dict.size(); ++i) {
    dict[i] = nextString();
    for (int j = 0; j < dict[i].size(); ++j) {
      dict[i][j] = tolower(dict[i][j]);
    }
  }
  memset(memo, -1, sizeof(memo));
  int res = solve("", 0, n);
  cout << res << endl;
  return 0;
}
