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 SD() {
  int i;
  scanf("%d", &i);
  return i;
}
long long SLL() {
  long long i;
  scanf("%lld", &i);
  return i;
}
const int AKA = 1e6 + 6;
char s[AKA];
int j, i, p, q, sz;
bool solve(int a, int b) {
  for (i = 1; i < 101; ++i)
    for (j = 1; j < 101; ++j)
      if (i * a + j * b == sz) return true;
  if (!(sz % p)) {
    i = sz / p, j = 0;
    return true;
  } else if (!(sz % q)) {
    j = sz / q, i = 0;
    return true;
  }
  return false;
}
int main() {
  int ACCEPTED = 1;
  while (ACCEPTED--) {
    sz = SD(), p = SD(), q = SD();
    scanf("%s", &s);
    if (!solve(p, q))
      puts("-1");
    else {
      printf("%d\n", i + j);
      int pnt = 0;
      for (int k = 0; k < i; k++) {
        for (int l = 0; l < p; ++l) printf("%c", s[pnt++]);
        puts("");
      }
      for (int k = 0; k < j; k++) {
        for (int l = 0; l < q; ++l) printf("%c", s[pnt++]);
        puts("");
      }
    }
  }
}
