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;
const long long INF = 1e18 + 1;
inline long long input(void) {
  char t;
  long long x = 0;
  int neg = 0;
  t = getchar();
  while ((t < 48 || t > 57) && t != '-') t = getchar();
  if (t == '-') {
    neg = 1;
    t = getchar();
  }
  while (t >= 48 && t <= 57) {
    x = (x << 3) + (x << 1) + t - 48;
    t = getchar();
  }
  if (neg) x = -x;
  return x;
}
inline void output(long long x) {
  char a[20];
  int i = 0, j;
  a[0] = '0';
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x == 0) putchar('0');
  while (x) {
    a[i++] = x % 10 + 48;
    x /= 10;
  }
  for (j = i - 1; j >= 0; j--) {
    putchar(a[j]);
  }
  putchar('\n');
}
int main() {
  long long n, p, q, flag = 0, k = 0;
  string s;
  n = input();
  p = input();
  q = input();
  cin >> s;
  for (int i = 0; i < n / q + 1; ++i) {
    if ((n - i * q) % p == 0) {
      k = i + (n - i * q) / p;
      output(k);
      for (int j = 0; j < i; ++j) {
        for (int w = j * q; w < (j + 1) * q; ++w) cout << s[w];
        cout << "\n";
      }
      for (int j = 0; j < (n - i * q) / p; ++j) {
        for (int w = i * q + j * p; w < i * q + (j + 1) * p; ++w) cout << s[w];
        cout << "\n";
      }
      flag = 1;
      break;
    }
  }
  if (!flag) printf("-1\n");
  return 0;
}
