"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;
int trie[TRIESIZE][26];
int trieSize;
int nextPref[TRIESIZE][26];
int maxWordInPref[TRIESIZE];
int prefLength[TRIESIZE];
int memo[TRIESIZE][WORDLENGTH + 1][LENGTH + 1];
int solve(int pref, int notCover, int length) {
  if (length == 0) {
    return notCover == 0;
  }
  int &res = memo[pref][notCover][length];
  if (res != -1) {
    return res;
  }
  res = 0;
  for (int c = 0; c < 26; ++c) {
    int npref = nextPref[pref][c];
    if (npref == 0) {
      continue;
    }
    int nnotCover = notCover + 1;
    if (maxWordInPref[npref] >= nnotCover) {
      nnotCover = 0;
    }
    if (nnotCover > prefLength[npref]) {
      continue;
    }
    res = (res + solve(npref, nnotCover, length - 1)) % MOD;
  }
  return res;
}
void initTrie() {
  memset(trie, -1, sizeof(trie));
  trieSize = 1;
  for (int i = 0; i < dict.size(); ++i) {
    int at = 0;
    for (int j = 0; j < dict[i].size(); ++j) {
      int c = dict[i][j] - 'a';
      if (trie[at][c] == -1) {
        prefLength[trieSize] = j + 1;
        trie[at][c] = trieSize++;
      }
      at = trie[at][c];
    }
  }
}
int getPref(string s) {
  int at = 0;
  for (int i = 0; i < s.size(); ++i) {
    int c = s[i] - 'a';
    if (trie[at][c] == -1) {
      return -1;
    }
    at = trie[at][c];
  }
  return at;
}
void initNextPref() {
  memset(nextPref, -1, sizeof(nextPref));
  for (int i = 0; i < dict.size(); ++i) {
    for (int prefLength = 0; prefLength <= dict[i].size(); ++prefLength) {
      string str = dict[i].substr(0, prefLength);
      int pref = getPref(str);
      for (int c = 0; c < 26; ++c) {
        if (nextPref[pref][c] == -1) {
          string ex = str;
          ex += c + 'a';
          for (int length = ex.size(); length >= 0; --length) {
            int npref = getPref(ex.substr(ex.size() - length));
            if (npref != -1) {
              nextPref[pref][c] = npref;
              break;
            }
          }
        }
      }
    }
  }
}
void initMaxWordInPref() {
  memset(maxWordInPref, -1, sizeof(maxWordInPref));
  for (int i = 0; i < dict.size(); ++i) {
    for (int prefLength = 0; prefLength <= dict[i].size(); ++prefLength) {
      string prefStr = dict[i].substr(0, prefLength);
      int pref = getPref(prefStr);
      if (maxWordInPref[pref] == -1) {
        for (int subLength = prefStr.size(); subLength >= 0; --subLength) {
          string subStr = prefStr.substr(prefStr.size() - subLength);
          for (int j = 0; j < dict.size(); ++j) {
            if (dict[j] == subStr) {
              maxWordInPref[pref] = subLength;
              break;
            }
          }
          if (maxWordInPref[pref] != -1) {
            break;
          }
        }
      }
    }
  }
}
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]);
    }
  }
  initTrie();
  initNextPref();
  initMaxWordInPref();
  memset(memo, -1, sizeof(memo));
  int res = solve(0, 0, n);
  cout << res << endl;
  return 0;
}
