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, a, b;
string s;
vector<int> ans;
int main() {
  scanf("%d%d%d", &n, &a, &b);
  cin >> s;
  s = " " + s;
  while (n > 0) {
    if (n % a == 0) {
      for (int i = 1; i <= n / a; ++i) ans.push_back(a);
      n = 0;
      break;
    }
    if (n % b == 0) {
      for (int i = 1; i <= n / b; ++i) ans.push_back(b);
      n = 0;
      break;
    }
    ans.push_back(a);
    ans.push_back(b);
    n -= a + b;
  }
  if (n == 0) {
    int sum = 0;
    cout << (int)ans.size() << endl;
    for (int i = 0; i < (int)ans.size(); ++i) {
      int now = ans[i];
      for (int j = sum + 1; j <= sum + ans[i]; ++j) cout << s[j];
      cout << endl;
      sum += ans[i];
    }
  } else
    printf("-1\n");
  return 0;
}
