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 exgcd(int a, int b, int &x, int &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  int r = exgcd(b, a % b, x, y);
  int t = x;
  x = y;
  y = t - (a / b) * y;
  return r;
}
int main(void) {
  int n, p, q, x, y;
  char ch[101];
  while (scanf("%d%d%d", &n, &p, &q) == 3) {
    scanf("%s", &ch);
    int d = exgcd(p, q, x, y);
    if (n % d != 0)
      printf("-1\n");
    else {
      int a = -n * x / q;
      int b = n * y / p;
      if (a > 0 && (n * x) % q != 0) a = a + 1;
      if (b < 0 && (n * y) % p != 0) b = b - 1;
      if (a > b)
        printf("-1\n");
      else {
        int t = b;
        int r = (n * x + q * t) / d;
        int l = (n * y - p * t) / d;
        printf("%d\n", r + l);
        for (int i = 0; i < r; i++) {
          for (int j = i * p; j < (i + 1) * p; j++) printf("%c", ch[j]);
          printf("\n");
        }
        for (int i = 0; i < l; i++) {
          for (int j = r * p + i * q; j < r * p + (i + 1) * q; j++)
            printf("%c", ch[j]);
          printf("\n");
        }
      }
    }
  }
  return 0;
}
