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 main() {
  int a, b, c;
  while (cin >> a >> b >> c) {
    char s[a];
    cin >> s;
    int k = 0, l = 1, m = 1, ans, f, y;
    int e = 0;
    bool g = false, d = false;
    if (a % b == 0) {
      ans = a / b;
      g = true;
      d = true;
      int e = 0;
      cout << ans << endl;
      for (int h = ans; h > 0; h--) {
        for (f = 0; f < b; f++) {
          cout << s[e];
          e++;
        }
        cout << endl;
      }
    } else if (a % c == 0) {
      ans = a / c;
      g = true;
      d = true;
      cout << ans << endl;
      for (int h = ans; h > 0; h--) {
        for (f = 0; f < c; f++) {
          cout << s[e];
          e++;
        }
        cout << endl;
      }
    } else {
      if (b > c) {
        int temp = b;
        b = c;
        c = temp;
      }
      for (int i = 1; c * m <= a; i++) {
        k = a - (b * i);
        if (k % c == 0) {
          e = 0;
          ans = (k / c) + i;
          cout << ans << endl;
          int p = 0;
          for (int j = 1; j <= i; j++) {
            for (int k = 0; k < b; k++) {
              cout << s[e];
              e++;
            }
            cout << endl;
          }
          for (int j = 1; j <= k / c; j++) {
            for (int k = 0; k < c; k++) {
              cout << s[e];
              e++;
            }
            cout << endl;
          }
          g = true;
          break;
        } else if (k < 1) {
          break;
        }
      }
    }
    if (g == false) {
      cout << -1 << endl;
    }
  }
}
