"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>
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long func[300];
vector<long long> hg;
struct karas {
  struct vert {
    long long nex[4];
    bool term;
    long long sz;
    long long p, p_c, link;
  };
  vector<vert> ms;
  long long cnt;
  karas() {
    ms.push_back(vert());
    cnt = 1;
  }
  void add(long long root, string &s, long long v) {
    if (v == s.size()) {
      ms[root].term = 1;
      ms[root].sz = s.size();
      return;
    }
    if (ms[root].nex[func[s[v]]] == 0) {
      ms[root].nex[func[s[v]]] = cnt++;
      ms.push_back(vert());
    }
    add(ms[root].nex[func[s[v]]], s, v + 1);
  }
  void build() {
    queue<long long> s;
    s.push(0);
    while (s.size()) {
      long long v = s.front();
      hg.push_back(v);
      s.pop();
      if (v == 0)
        ms[v].link = 0;
      else {
        ms[v].link = ms[ms[ms[v].p].link].nex[ms[v].p_c];
        if (ms[v].link == v) ms[v].link = 0;
      }
      if (!ms[v].term) {
        ms[v].term = ms[ms[v].link].term;
        ms[v].sz = ms[ms[v].link].sz;
      }
      for (long long i = 0; i < 4; i++) {
        if (ms[v].nex[i] != 0) {
          ms[ms[v].nex[i]].p = v;
          ms[ms[v].nex[i]].p_c = i;
          s.push(ms[v].nex[i]);
        } else
          ms[v].nex[i] = ms[ms[v].link].nex[i];
      }
    }
  }
};
long long dp[1113][12][105];
const long long mod = 1e9 + 9;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.setf(ios::fixed);
  cout.precision(0);
  long long n, m;
  cin >> n >> m;
  karas ms;
  func['A'] = 0, func['C'] = 1, func['G'] = 2, func['T'] = 3;
  for (long long i = 0; i < m; i++) {
    string s;
    cin >> s;
    ms.add(0, s, 0);
  }
  ms.build();
  dp[0][0][0] = 1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < 11; j++)
      for (long long k1 = hg.size() - 1; k1 >= 0; k1--) {
        long long k = hg[k1];
        bool sd = false;
        for (long long e = 0; e < 4; e++) {
          long long nd = ms.ms[k].nex[e];
          if (ms.ms[nd].term && ms.ms[nd].sz >= j + 1)
            dp[i + 1][0][nd] = (dp[i + 1][0][nd] + dp[i][j][k]) % mod;
          else
            dp[i + 1][j + 1][nd] = (dp[i + 1][j + 1][nd] + dp[i][j][k]) % mod;
        }
      }
  }
  long long ans = 0;
  for (long long k = 0; k < ms.ms.size(); k++) ans = (ans + dp[n][0][k]) % mod;
  cout << ans;
}
