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 main() {
  int n, p, q, ans, i, j;
  string str;
  while (scanf("%d%d%d\n", &n, &p, &q) != EOF) {
    int a = -1, b = -1;
    cin >> str;
    for (i = 0; i <= n; i++) {
      for (j = 0; j <= n; j++) {
        ans = i * p + j * q;
        if (ans == n) {
          a = i;
          b = j;
          break;
        }
      }
    }
    if (a == -1 || b == -1)
      cout << "-1" << endl;
    else {
      int temp1, temp2, temp3, k;
      temp1 = a * p;
      temp2 = b * q;
      temp3 = a + b;
      cout << temp3 << endl;
      int count = 0;
      for (i = 0; i < temp1; i++) {
        cout << str[i];
        count++;
        if (count == p) cout << endl, count = 0;
      }
      count = 0;
      for (i = temp1; i < temp1 + temp2; i++) {
        cout << str[i];
        ;
        count++;
        if (count == q) cout << endl, count = 0;
      }
    }
  }
}
