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 T = 1, c = 0;
  while (T--) {
    vector<long long> v, v1;
    long long n, m, j, y, i, k, x, d, l, p, r;
    cin >> n >> x >> y;
    string s;
    cin >> s;
    if (n % x != 0 && n % y != 0 && x + y > n) {
      cout << -1 << endl;
      continue;
    }
    if (n % x == 0) {
      cout << n / x << endl;
      for (i = 0; i < s.size(); i++) {
        for (j = i; j < i + x && i + x <= s.size(); j++) {
          cout << s[j];
        }
        i = j - 1;
        cout << endl;
      }
    } else if (n % y == 0) {
      cout << n / y << endl;
      for (i = 0; i < s.size(); i++) {
        for (j = i; j < i + y && i + y <= s.size(); j++) {
          cout << s[j];
        }
        i = j - 1;
        cout << endl;
      }
    } else {
      bool f = false;
      long long mx = max(x, y);
      long long mn = min(x, y);
      l = 0;
      r = 0;
      for (i = 1;; i++) {
        if ((n - (mx * i)) % mn == 0 && i * mx <= n) {
          f = true;
          l = i;
          r = (n - (mx * i)) / mn;
          break;
        } else if (mx * i > n)
          break;
      }
      if (f) {
        cout << l + r << endl;
        long long c = 0, z = 0;
        for (i = 0; i < s.size(); i++) {
          if (c == l) {
            z = i;
            break;
          }
          for (j = i; j < i + mx && i + mx <= s.size(); j++) {
            cout << s[j];
          }
          c++;
          i = j - 1;
          cout << endl;
        }
        c = 0;
        for (i = z; i < s.size(); i++) {
          if (c == r) break;
          for (j = i; j < i + mn && i + mn <= s.size(); j++) {
            cout << s[j];
          }
          c++;
          i = j - 1;
          cout << endl;
        }
      } else
        cout << -1 << endl;
    }
  }
  return 0;
}
