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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
double startTime;
double gct() { return ((double)clock() - startTime) / CLOCKS_PER_SEC; }
void suraj() {
  int n, a, b;
  cin >> n >> a >> b;
  string s;
  cin >> s;
  if (n % a == 0) {
    cout << n / a << "\n";
    int c = n / a;
    for (int i = 0, j = 0; j < c; j++, i += a) cout << s.substr(i, a) << "\n";
  } else if (n % b == 0) {
    cout << n / b << "\n";
    int c = n / b;
    for (int i = 0, j = 0; j < c; j++, i += b) cout << s.substr(i, b) << "\n";
  } else if (n % (a + b) == 0) {
    cout << 2 * n / (a + b) << "\n";
    int c = n / (a + b);
    int i = 0;
    int j = 0;
    for (i = 0, j = 0; j < c; i += a, j++) cout << s.substr(i, a) << "\n";
    for (j = 0; j < c; j++, i += b) cout << s.substr(i, b) << "\n";
  } else {
    for (int i = 0; i < 101; i++) {
      if ((n - a * i > 0) && (n - a * i) % b == 0) {
        int aa = i;
        int bb = (n - i * a) / b;
        int ptr = 0;
        cout << aa + bb << "\n";
        for (int k = 0; k < aa; k++) {
          cout << s.substr(ptr, a) << "\n";
          ptr += a;
        }
        for (int k = 0; k < bb; k++) {
          cout << s.substr(ptr, b) << "\n";
          ptr += b;
        }
        return;
      } else if ((n - b * i) > 0 && (n - b * i) % a == 0) {
        int bb = i;
        int aa = (n - i * b) / a;
        int ptr = 0;
        cout << aa + bb << "\n";
        for (int k = 0; k < bb; k++) {
          cout << s.substr(ptr, b) << "\n";
          ptr += b;
        }
        for (int k = 0; k < aa; k++) {
          cout << s.substr(ptr, a) << "\n";
          ptr += a;
        }
        return;
      }
    }
    cout << -1 << "\n";
  }
}
int main() {
  startTime = (double)clock();
  suraj();
  return 0;
}
