"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;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
const int MOD = 1e9 + 9;
const int MAXN = 1e3 + 7;
char buf[MAXN];
int n, m;
void add(int &x, int y) { x = x + y >= MOD ? x + y - MOD : x + y; }
struct Ac_Automaton {
  int ch[MAXN][26], fail[MAXN], tot, len[MAXN];
  int newnode() {
    tot++;
    memset(ch[tot], 0, sizeof(ch[tot]));
    fail[tot] = len[tot] = 0;
    return tot;
  }
  Ac_Automaton() {
    tot = -1;
    newnode();
  }
  void insert(char *s) {
    int u = 0;
    for (int i = 0; s[i]; i++) {
      int c = s[i] - 'A';
      if (!ch[u][c]) ch[u][c] = newnode();
      u = ch[u][c];
    }
    len[u] = strlen(s);
  }
  void buildFail() {
    queue<int> que;
    for (int c = 0; c < 26; c++)
      if (ch[0][c]) {
        que.push(ch[0][c]);
        fail[ch[0][c]] = 0;
      }
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      len[u] = max(len[u], len[fail[u]]);
      for (int c = 0; c < 26; c++) {
        if (!ch[u][c]) {
          ch[u][c] = ch[fail[u]][c];
          continue;
        }
        que.push(ch[u][c]);
        int v = fail[u];
        while (v and !ch[v][c]) v = fail[v];
        fail[ch[u][c]] = ch[v][c];
      }
    }
  }
  int work(int n) {
    vector<vector<int> > f(tot + 1, vector<int>(11, 0));
    f[0][0] = 1;
    for (int t = 1; t <= n; t++) {
      vector<vector<int> > next_f(tot + 1, vector<int>(11, 0));
      for (int i = 0; i <= tot; i++) {
        for (int mat = 0; mat < 10; mat++) {
          for (int c = 0; c < 26; c++) {
            int nextp = ch[i][c];
            if (len[nextp] >= mat + 1)
              add(next_f[nextp][0], f[i][mat]);
            else
              add(next_f[nextp][mat + 1], f[i][mat]);
          }
        }
      }
      f.swap(next_f);
    }
    int ret = 0;
    for (int i = 0; i <= tot; i++) add(ret, f[i][0]);
    return ret;
  }
} aho;
void solve() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%s", buf + 1);
    aho.insert(buf + 1);
  }
  aho.buildFail();
  cout << aho.work(n) << endl;
}
int main() {
  solve();
  return 0;
}
