You are given the string s of length n and the numbers p, q. Split the string s to pieces of length p and q.

For example, the string "Hello" for p = 2, q = 3 can be split to the two strings "Hel" and "lo" or to the two strings "He" and "llo".

Note it is allowed to split the string s to the strings only of length p or to the strings only of length q (see the second sample test).

Input

The first line contains three positive integers n, p, q (1 ≤ p, q ≤ n ≤ 100).

The second line contains the string s consists of lowercase and uppercase latin letters and digits.

Output

If it's impossible to split the string s to the strings of length p and q print the only number "-1".

Otherwise in the first line print integer k — the number of strings in partition of s.

Each of the next k lines should contain the strings in partition. Each string should be of the length p or q. The string should be in order of their appearing in string s — from left to right.

If there are several solutions print any of them.

Examples

Input

5 2 3
Hello


Output

2
He
llo


Input

10 9 5
Codeforces


Output

2
Codef
orces


Input

6 4 5
Privet


Output

-1


Input

8 1 1
abacabac


Output

8
a
b
a
c
a
b
a
c
#include <bits/stdc++.h>
using namespace std;
int dp[200][200];
int sol[200][200][2];
int main() {
  string s;
  int n, p, q;
  cin >> n >> p >> q;
  cin >> s;
  int k = 0;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int sum = 0; sum <= n; sum++) {
      if (dp[i - 1][sum] == 1) {
        dp[i][sum] = 1;
        sol[i][sum][0] = sol[i - 1][sum][0];
        sol[i][sum][1] = sol[i - 1][sum][1];
        if (sum + p <= n) {
          if (dp[i][sum + p] == 0) {
            dp[i][sum + p] = 1;
            sol[i][sum + p][0] = sol[i - 1][sum][0];
            sol[i][sum + p][1] = sol[i - 1][sum][1];
            sol[i][sum + p][0]++;
          }
        }
        if (sum + q <= n) {
          if (dp[i][sum + q] == 0) {
            dp[i][sum + q] = 1;
            sol[i][sum + q][0] = sol[i - 1][sum][0];
            sol[i][sum + q][1] = sol[i - 1][sum][1];
            sol[i][sum + q][1]++;
          }
        }
      }
    }
    int first = 0;
    if (dp[i][n] == 1) {
      cout << sol[i][n][0] + sol[i][n][1] << endl;
      for (int l = 0; l < sol[i][n][0]; l++) {
        for (int j = first; j < first + p; j++) {
          cout << s[j];
        }
        first += p;
        cout << endl;
      }
      for (int l = 0; l < sol[i][n][1]; l++) {
        for (int j = first; j < first + q; j++) {
          cout << s[j];
        }
        first += q;
        cout << endl;
      }
      return 0;
    }
  }
  cout << -1 << endl;
}
