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>
int main() {
  int a, b, c, i, j, k = -1, n = 0, t, w;
  scanf("%d%d%d", &a, &b, &c);
  char str[a + 1];
  scanf("%s", str);
  if (b + c == a) {
    puts("2");
    for (i = 0; i < b; i++) {
      printf("%c", str[i]);
    }
    printf("\n");
    for (i = b; i < a; i++) {
      printf("%c", str[i]);
    }
  } else if (a % c == 0) {
    printf("%d\n", a / c);
    for (i = 1; i <= a / c; i++) {
      for (j = 0; j < c; j++) {
        k++;
        printf("%c", str[k]);
      }
      printf("\n");
    }
  } else if (a % b == 0) {
    printf("%d\n", a / b);
    for (i = 1; i <= a / b; i++) {
      for (j = 0; j < b; j++) {
        k++;
        printf("%c", str[k]);
      }
      printf("\n");
    }
  } else {
    i = a;
    while (i % c != 0 && i >= 0) {
      i = i - b;
      n++;
    }
    if (i % c == 0 && i >= 0) {
      t = i / c;
      printf("%d\n", t + n);
      for (i = 1; i <= n; i++) {
        for (j = 0; j < b; j++) {
          k++;
          printf("%c", str[k]);
        }
        printf("\n");
      }
      for (i = 1; i <= t; i++) {
        for (j = 0; j < c; j++) {
          k++;
          printf("%c", str[k]);
        }
        printf("\n");
      }
    } else if (i <= 0) {
      puts("-1");
    }
  }
}
