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 = 50 + 9;
const long long inf = 1e8 + 2;
int Extgcd(int a, int b, int& x, int& y);
int main() {
  int n, p, q;
  while (~scanf("%d%d%d", &n, &p, &q)) {
    char aha[200];
    scanf("%s", aha);
    int x, y;
    int gg = Extgcd(p, q, x, y);
    if (n % gg != 0) {
      printf("-1\n");
      continue;
    }
    x *= n / gg, y *= n / gg;
    int tmp = q / gg;
    x = (x % tmp + tmp) % tmp;
    y = n - p * x;
    y /= q;
    if (y < 0) {
      printf("-1\n");
      continue;
    }
    int loca = 0;
    printf("%d\n", x + y);
    for (int i = 0; i < x; i++) {
      for (int j = 0; j < p; j++) printf("%c", aha[loca++]);
      printf("\n");
    }
    for (int i = 0; i < y; i++) {
      for (int j = 0; j < q; j++) printf("%c", aha[loca++]);
      printf("\n");
    }
  }
  return 0;
}
int Extgcd(int a, int b, int& x, int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    int ans = Extgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - (a / b) * y;
    return ans;
  }
}
