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 dp[1000][1000];
int main() {
  int n, p, q;
  bool cam = false;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  if (p > q) {
    swap(p, q);
    cam = true;
  }
  if (p + q > n && n % q != 0 && n % p != 0) {
    cout << -1 << endl;
  } else if (p == q && n % p == 0) {
    cout << n / q << endl;
    for (int i = 0; i < n; i += q) cout << s.substr(i, q) << endl;
  } else if (p + q == n) {
    cout << 2 << endl;
    if (!cam)
      cout << s.substr(0, p) << endl << s.substr(p, n) << endl;
    else
      cout << s.substr(0, q) << endl << s.substr(q, n) << endl;
  } else if (n % p == 0) {
    cout << n / p << endl;
    for (int i = 0; i < n; i += p) cout << s.substr(i, p) << endl;
  } else if (n % q == 0) {
    cout << n / q << endl;
    for (int i = 0; i < n; i += q) cout << s.substr(i, q) << endl;
  }
  if (p + q < n && n % q != 0 && n % p != 0) {
    int suma = 0, i = n;
    while (i > 0 && i % p != 0) {
      i -= q;
      suma++;
    }
    if (i > 0) {
      cout << suma + i / p << endl;
      for (int j = 0; j < q * suma; j += q) cout << s.substr(j, q) << endl;
      for (int j = q * suma; j < n; j += p) {
        cout << s.substr(j, p) << endl;
      }
    } else {
      cout << -1 << endl;
    }
  }
  return 0;
}
