"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 = 0x7FFFFFFF;
const long long MOD = 1e9 + 9;
struct AhoCorasickAutomata {
  static const int CHARSET_SIZE = 4;
  static const int MAX_WORD_CNT = 15;
  static const int MAX_WORD_LEN = 15;
  static const int PREALLOCATE_SIZE = MAX_WORD_CNT * MAX_WORD_LEN + 10;
  static const int TRIE_ROOT = 1;
  typedef struct _node {
    int ch[CHARSET_SIZE];
    int fail;
    int mxMth;
  } node, *pnode;
  node preAlloc[PREALLOCATE_SIZE];
  int AllocCnt;
  inline int getID(char c) {
    switch (c) {
      case 'A':
        return 0;
      case 'T':
        return 1;
      case 'C':
        return 2;
      default:
        return 3;
    }
  }
  inline void AllocNew() {
    AllocCnt++;
    for (int i = 0; i < CHARSET_SIZE; i++) preAlloc[AllocCnt].ch[i] = 0;
    preAlloc[AllocCnt].fail = 0;
    preAlloc[AllocCnt].mxMth = 0;
  }
  void clear() {
    AllocCnt = 0;
    AllocNew();
  }
  void insert(const char* s) {
    int len = strlen(s);
    int p = TRIE_ROOT, q = 0;
    for (int i = 0; i < len; i++) {
      int t = getID(s[i]);
      q = preAlloc[p].ch[t];
      if (q < TRIE_ROOT) {
        AllocNew();
        q = AllocCnt;
        preAlloc[p].ch[t] = q;
      }
      p = q;
    }
    preAlloc[p].mxMth = len;
  }
  void initFail() {
    queue<int> q;
    int p(TRIE_ROOT);
    preAlloc[p].fail = p;
    for (int i = 0; i < CHARSET_SIZE; i++)
      if (preAlloc[TRIE_ROOT].ch[i] > TRIE_ROOT)
        preAlloc[preAlloc[TRIE_ROOT].ch[i]].fail = TRIE_ROOT,
        q.push(preAlloc[TRIE_ROOT].ch[i]);
      else
        preAlloc[TRIE_ROOT].ch[i] = TRIE_ROOT;
    while (!q.empty()) {
      p = q.front();
      q.pop();
      preAlloc[p].mxMth =
          max(preAlloc[p].mxMth, preAlloc[preAlloc[p].fail].mxMth);
      for (int i = 0; i < CHARSET_SIZE; i++)
        if (preAlloc[p].ch[i] < TRIE_ROOT)
          preAlloc[p].ch[i] = preAlloc[preAlloc[p].fail].ch[i];
        else
          preAlloc[preAlloc[p].ch[i]].fail = preAlloc[preAlloc[p].fail].ch[i],
          q.push(preAlloc[p].ch[i]);
    }
  }
  AhoCorasickAutomata() { this->clear(); }
} Main;
char op[15];
int N, M;
long long dp[1010][110][15];
int solve(const AhoCorasickAutomata& ACA) {
  memset(dp, 0x00, sizeof dp);
  dp[0][1][0] = 1;
  for (int i = 0; i < N; i++)
    for (int j = 1; j <= ACA.AllocCnt; j++)
      for (int k = 0; k <= 10; k++)
        for (int c = 0; c < AhoCorasickAutomata::CHARSET_SIZE; c++) {
          int t = ACA.preAlloc[j].ch[c];
          if (ACA.preAlloc[t].mxMth >= k + 1)
            dp[i + 1][t][0] = (dp[i + 1][t][0] + dp[i][j][k]) % MOD;
          else
            dp[i + 1][t][k + 1] = (dp[i + 1][t][k + 1] + dp[i][j][k]) % MOD;
        }
  long long ans = 0;
  for (int i = 1; i <= ACA.AllocCnt; i++) ans = (ans + dp[N][i][0]) % MOD;
  return ans;
}
int main() {
  cin >> N >> M;
  for (int i = 1; i <= M; i++) cin >> op, Main.insert(op);
  Main.initFail();
  cout << solve(Main) << endl;
  return 0;
}
