Problem:
Score : 500 points

Problem Statement

Takahashi has decided to work on K days of his choice from the N days starting with tomorrow.

You are given an integer C and a string S. Takahashi will choose his workdays as follows:

After working for a day, he will refrain from working on the subsequent C days.

If the i-th character of S is x, he will not work on Day i, where Day 1 is tomorrow, Day 2 is the day after tomorrow, and so on.

Find all days on which Takahashi is bound to work.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq N

0 \leq C \leq N

The length of S is N.

Each character of S is o or x.

Takahashi can choose his workdays so that the conditions in Problem Statement are satisfied.

Input

Input is given from Standard Input in the following format:

N K C
S

Output

Print all days on which Takahashi is bound to work in ascending order, one per line.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

int main(){

	int n, k, c, aa, bb;
	string s;
	vector<int> a, b;

	cin>> n >> k >> c;
	cin>> s;

	for (int i = 0;i < n;i++) {
		if (s[i] == 'o') {
			a.push_back(i + 1);
			i += c - 1;
		}
		if (a.size() == k) break;
	}

	for (int i = n-1;i>=0;i--) {
		if (s[i] == 'o') {
			b.push_back(i + 1);
			i -= c - 1;
		}
		if (b.size() == k) break;
	}

	sort(a.begin(), a.end());
	sort(b.begin(), b.end());

	aa = 0;
	bb = 0;
	while (1) {
		if (a[aa] == b[bb]) cout << a[aa] << endl;
		else if (a[aa] < b[bb]) bb--;
		else aa--;
		aa++;
		bb++;

		if (aa == k || bb == k) break;
	}

	return 0;
}