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 s, p, q;
  char str[101];
  while (scanf("%d%d%d", &s, &p, &q) != EOF) {
    memset(str, 0, sizeof(str));
    scanf("%s", str);
    int x, y;
    x = s / p;
    y = s / q;
    int a = 0, b = 0;
    int find_ans = 0;
    for (int i = 0; i <= x; i++) {
      int sum = 0;
      if (!find_ans) {
        for (int j = 0; j <= y; j++) {
          sum = i * p + j * q;
          if (sum == s) {
            a = i;
            b = j;
            find_ans = 1;
            break;
          }
        }
      } else
        break;
    }
    if (find_ans) {
      printf("%d\n", a + b);
      int u = 0;
      for (int i = 0; i < a; i++) {
        for (int j = 0; j < p; j++)
          if (j != p - 1) {
            printf("%c", str[u + j]);
          } else {
            printf("%c\n", str[u + j]);
          }
        u += p;
      }
      for (int i = 0; i < b; i++) {
        for (int j = 0; j < q; j++)
          if (j != q - 1) {
            printf("%c", str[u + j]);
          } else {
            printf("%c\n", str[u + j]);
          }
        u += q;
      }
    } else
      printf("-1\n");
  }
  return 0;
}
