"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 PMA {
  PMA *next[0x100];
  vector<int> accept;
  PMA() { fill(next, next + 0x100, (PMA *)0); }
};
PMA *buildPMA(char p[10][11], int size) {
  PMA *root = new PMA;
  for (int i = 0; i < size; ++i) {
    PMA *t = root;
    for (int j = 0; p[i][j]; ++j) {
      char c = p[i][j];
      if (t->next[c] == NULL) t->next[c] = new PMA;
      t = t->next[c];
    }
    t->accept.push_back(i);
  }
  queue<PMA *> Q;
  for (int c = 'A'; c <= 'Z'; ++c) {
    if (root->next[c]) {
      root->next[c]->next[0] = root;
      Q.push(root->next[c]);
    } else
      root->next[c] = root;
  }
  while (!Q.empty()) {
    PMA *t = Q.front();
    Q.pop();
    for (int c = 'A'; c <= 'Z'; ++c) {
      if (t->next[c]) {
        Q.push(t->next[c]);
        PMA *r = t->next[0];
        while (!r->next[c]) r = r->next[0];
        t->next[c]->next[0] = r->next[c];
        for (__typeof(r->next[c]->accept.begin()) i =
                 r->next[c]->accept.begin();
             i != r->next[c]->accept.end(); i++)
          t->next[c]->accept.push_back(*i);
      }
    }
  }
  return root;
}
int match(char *t, PMA *v, int *result) {
  int n = strlen(t);
  int count = 0;
  for (int i = 0; i < n; ++i) {
    char c = t[i];
    while (!v->next[c]) v = v->next[0];
    v = v->next[c];
    for (int j = 0; j < v->accept.size(); ++j) result[v->accept[j]]++;
  }
}
const int mod = (int)1e9 + 9;
int n, m;
map<pair<int, PMA *>, int> dp[2];
int len[10];
char in[10][11];
const char *base = "ACGT";
void run() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> in[i];
    len[i] = strlen(in[i]);
  }
  PMA *pma = buildPMA(in, m);
  dp[0][make_pair(0, pma)] = 1;
  int cur = 0, next = 1;
  for (int i = 0; i < n; i++) {
    dp[next].clear();
    for (__typeof(dp[cur].begin()) it = dp[cur].begin(); it != dp[cur].end();
         it++) {
      for (int b = 0; b < 4; b++) {
        PMA *v = it->first.second;
        while (!v->next[base[b]]) v = v->next[0];
        v = v->next[base[b]];
        int mxlen = -1;
        for (__typeof(v->accept.begin()) jt = v->accept.begin();
             jt != v->accept.end(); jt++)
          mxlen = max(mxlen, len[*jt]);
        if (it->first.first < mxlen)
          (dp[next][make_pair(0, v)] += it->second) %= mod;
        else if (it->first.first + 1 < 12)
          (dp[next][make_pair(it->first.first + 1, v)] += it->second) %= mod;
      }
    }
    swap(cur, next);
  }
  int ans = 0;
  for (__typeof(dp[cur].begin()) i = dp[cur].begin(); i != dp[cur].end(); i++)
    if (i->first.first == 0) ans = (ans + i->second) % mod;
  cout << ans << endl;
}
int main() {
  run();
  return 0;
}
