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 n, p, q, j, k;
string s;
int main() {
  cin >> n >> p >> q >> s;
  int cnt1 = 0, cnt2 = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      if (i * p + j * q == n) {
        cnt1 = i;
        cnt2 = j;
        goto go;
      }
    }
  }
go:
  if (q + p == n) {
    cout << 2 << endl;
    for (int i = 0; i < n - q; i++) cout << s[i];
    cout << endl;
    for (int i = n - q; i < n; i++) cout << s[i];
  } else if (n % p == 0) {
    cout << n / p << endl;
    int m = n / p;
    int i = 0;
    while (m--) {
      for (int i = k; i < n / (n / p) + k; i++) cout << s[i], j = i + 1;
      k = j;
      cout << endl;
    }
  } else if (n % q == 0) {
    cout << n / q << endl;
    int m = n / q;
    while (m--) {
      for (int i = k; i < n / (n / q) + k; i++) cout << s[i], j = i + 1;
      k = j;
      cout << endl;
    }
  } else if (cnt1 || cnt2) {
    cout << cnt1 + cnt2 << endl;
    int m = cnt1;
    while (m--) {
      for (int i = k; i < p + k; i++) cout << s[i], j = i + 1;
      k = j;
      cout << endl;
    }
    while (cnt2--) {
      for (int i = k; i < q + k; i++) cout << s[i], j = i + 1;
      k = j;
      cout << endl;
    }
  } else
    cout << -1;
}
