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;
const int maxn = 1e5 + 5;
const int inf = 1e9 + 7;
int n, p, q;
char s[maxn];
int main() {
  while (scanf("%d%d%d", &n, &p, &q) != EOF) {
    scanf("%s", s);
    int x, y;
    bool ans = false;
    for (int i = 0; i <= 100; i++) {
      x = i;
      y = (n - p * x) / q;
      if (y >= 0 && x * p + y * q == n) {
        ans = true;
        break;
      }
    }
    if (!ans) {
      printf("-1\n");
    } else {
      printf("%d\n", x + y);
      int now = 0;
      for (int i = 1; i <= x; i++) {
        for (int j = 1; j <= p; j++) printf("%c", s[now++]);
        puts("");
      }
      for (int i = 1; i <= y; i++) {
        for (int j = 1; j <= q; j++) printf("%c", s[now++]);
        puts("");
      }
    }
  }
  return 0;
}
