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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100;
int n, p, q;
char s[MAXN + 1];
void run() {
  scanf("%d%d%d%s", &n, &p, &q, s);
  for (int i = 0; i * p <= n; ++i)
    if ((n - i * p) % q == 0) {
      int j = (n - i * p) / q, at = 0;
      printf("%d\n", i + j);
      for (int k = (0); k < (i); ++k) {
        for (int x = (0); x < (p); ++x) printf("%c", s[at++]);
        puts("");
      }
      for (int k = (0); k < (j); ++k) {
        for (int x = (0); x < (q); ++x) printf("%c", s[at++]);
        puts("");
      }
      return;
    }
  printf("-1\n");
}
int main() {
  run();
  return 0;
}
