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 int maxx = 103;
int dp[maxx], dad[maxx], n, p, q, sz[maxx];
string s;
vector<string> v;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> p >> q;
  cin >> s;
  sz[0] = 0;
  dp[0] = 1;
  dad[0] = -1;
  for (int i = 1; i <= n; i++) {
    if (i >= p && dp[i - p])
      dp[i] = 1, sz[i] = sz[i - p] + 1, dad[i] = i - p;
    else if (i >= q && dp[i - q])
      dp[i] = 1, sz[i] = sz[i - q] + 1, dad[i] = i - q;
  }
  if (!dp[n]) return cout << -1 << endl, 0;
  cout << sz[n] << endl;
  while (n != -1) {
    if (n) v.push_back(s.substr(dad[n], n - dad[n]));
    n = dad[n];
  }
  reverse(v.begin(), v.end());
  for (string x : v) cout << x << endl;
  return 0;
}
