"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 = 1e9 + 7, N = 1e3 + 7, A = 4, M = 1e9 + 9;
int idx(char c) {
  if (c == 'A')
    return 0;
  else if (c == 'C')
    return 1;
  else if (c == 'G')
    return 2;
  else
    return 3;
}
void add(int &a, int b) {
  a += b;
  if (a >= M) a -= M;
}
struct AC {
  int nxt[N][A], fail[N], yes[N], depth[N];
  int root, cnt;
  AC() {
    memset(nxt, 0, sizeof nxt);
    memset(fail, 0, sizeof fail);
    memset(yes, 0, sizeof yes);
    root = 0;
    cnt = 1;
    depth[root] = 0;
  }
  void insert(string s) {
    int u = root;
    for (char c : s) {
      int cc = idx(c);
      if (!nxt[u][cc]) nxt[u][cc] = cnt++;
      depth[nxt[u][cc]] = depth[u] + 1;
      u = nxt[u][cc];
    }
    yes[u] = 1;
  }
  void build_fail() {
    queue<int> q;
    for (int c = 0; c < A; c++) {
      if (nxt[0][c]) {
        fail[nxt[0][c]] = 0;
        q.push(nxt[0][c]);
      }
    }
    while (q.size()) {
      int u = q.front();
      q.pop();
      for (int c = 0; c < A; c++) {
        if (nxt[u][c]) {
          int v = nxt[u][c];
          q.push(v);
          int r = fail[u];
          while (r && !nxt[r][c]) r = fail[r];
          fail[v] = nxt[r][c];
        }
      }
    }
  }
  int dp[N][N][12];
  void solve(int n) {
    memset(dp, 0, sizeof dp);
    dp[0][0][0] = 1;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < cnt; j++) {
        for (int len = 0; len <= 10; len++) {
          if (!dp[i][j][len]) continue;
          for (int c = 0; c < A; c++) {
            int r = j;
            while (r && !nxt[r][c]) r = fail[r];
            if (!nxt[r][c]) continue;
            if (depth[r] < len) continue;
            r = nxt[r][c];
            int nwlen = len + 1;
            int rr = r;
            while (rr) {
              if (depth[rr] < len + 1) break;
              if (yes[rr]) {
                nwlen = 0;
                break;
              }
              rr = fail[rr];
            }
            add(dp[i + 1][r][nwlen], dp[i][j][len]);
          }
        }
      }
    }
    int ans = 0;
    for (int i = 0; i < cnt; i++) add(ans, dp[n][i][0]);
    cout << ans << endl;
  }
} ac;
signed main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    string s;
    cin >> s;
    ac.insert(s);
  }
  ac.build_fail();
  ac.solve(n);
}
