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;
inline long long int add(long long int a, long long int b) {
  long long int c = (a % 1000000007) + (b % 1000000007);
  if (c >= 1000000007) c -= 1000000007;
  return c;
}
inline long long int mul(long long int a, long long int b) {
  return ((a % 1000000007) * 1ll * (b % 1000000007)) % 1000000007;
}
inline long long int sub(long long int a, long long int b) {
  long long int c = (a % 1000000007) - (b % 1000000007);
  if (c < 0) c += 1000000007;
  return c;
}
const long long int nax = 300 * 1000 + 10;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void test_case();
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t = 1;
  while (t--) {
    test_case();
  }
  return 0;
}
void test_case() {
  long long int n, p, q;
  cin >> n >> p >> q;
  string a;
  cin >> a;
  for (long long int i = 0; i <= n; ++i) {
    for (long long int j = 0; j <= n; ++j) {
      if (i * p + j * q == n) {
        cout << i + j << '\n';
        string x = a.substr(0, i * p);
        string y = a.substr(i * p);
        string cur;
        for (char ch : x) {
          cur.push_back(ch);
          if ((long long int)cur.size() == p) {
            cout << cur << '\n';
            cur.clear();
          }
        }
        for (char ch : y) {
          cur.push_back(ch);
          if ((long long int)cur.size() == q) {
            cout << cur << '\n';
            cur.clear();
          }
        }
        return;
      }
    }
  }
  cout << -1;
}
