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;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << "[" << H << "]";
  debug_out(T...);
}
clock_t startTime;
double getCurrentTime() {
  return (double)(clock() - startTime) / CLOCKS_PER_SEC;
}
const int MOD = 1e9 + 7;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= MOD;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % MOD;
    a = a * a % MOD;
  }
  return res;
}
template <typename t1, typename t2>
inline bool upmax(t1 &a, t2 b) {
  if (a < (t1)b) {
    a = (t1)b;
    return true;
  } else
    return false;
}
template <typename t1, typename t2>
inline bool upmin(t1 &a, t2 b) {
  if (a > (t1)b) {
    a = (t1)b;
    return true;
  } else
    return false;
}
void solve(int TC) {
  int n, p, q;
  cin >> n >> p >> q;
  string s;
  cin >> s;
  for (int i = 0; i <= 200; i++) {
    for (int j = 0; j <= 200; j++) {
      if (i * p + q * j == n) {
        cout << i + j << endl;
        for (int k = 0; k < i; k++) {
          cout << s.substr(k * p, p) << endl;
        }
        for (int k = 0; k < j; k++) {
          cout << s.substr(p * i + k * q, q) << endl;
        }
        return;
      }
    }
  }
  cout << -1 << endl;
}
int main() {
  startTime = clock();
  int T = 1;
  for (int i = 1; i <= T; i++) {
    cerr << "Case #" << i << ":" << endl;
    solve(T);
  }
  cerr << "Time: " << getCurrentTime() << " ms" << endl;
  return 0;
}
