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 n, p, q;
  scanf("%d %d %d", &n, &p, &q);
  char word[n + 1];
  scanf("%s", word);
  if (n == p + q) {
    printf("2\n");
    for (int i = 0; i < p; ++i) {
      printf("%c", word[i]);
    }
    printf("\n");
    for (int i = p; i < n; ++i) {
      printf("%c", word[i]);
    }
    printf("\n");
    return 0;
  } else if (n % p != 0 && n % q != 0) {
    int z = 0, i;
    for (i = 0; i < n; ++i) {
      int x = n - (p * i);
      if (x % q == 0 && x > 0) {
        z = i;
        break;
      }
    }
    if (z == 0) {
      printf("-1\n");
      return 0;
    }
    int y = (n - (z * p)) / q;
    printf("%d\n", z + y);
    for (i = 0; i < z; ++i) {
      for (int j = 0; j < p; ++j) {
        int f = p * i + j;
        printf("%c", word[f]);
      }
      printf("\n");
    }
    for (i = 0; i < y; ++i) {
      int f = z * p;
      for (int j = 0; j < q; ++j) {
        int g = f + q * i + j;
        printf("%c", word[g]);
      }
      printf("\n");
    }
    return 0;
  } else {
    int s;
    if (n % p == 0) {
      s = p;
    } else {
      s = q;
    }
    printf("%d\n", n / s);
    for (int i = 0; i < n / s; ++i) {
      for (int j = 0; j < s; ++j) {
        int x = i * s + j;
        printf("%c", word[x]);
      }
      printf("\n");
    }
  }
  return 0;
}
