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;
#define rep(i, n) for (int i = 0; i < n; i++)

int main()
{
	int n, k, c;
	cin >> n >> k >> c;
	string str;
	cin >> str;
	vector<int> hataraku_L(n, 0);
	vector<int> hataraku_R(n, 0);
	vector<vector<bool>> ans(n, vector<bool>(2, false));
	vector<int> final_ans;
	int temp1 = 0;
	int temp2 = n - 1;
	rep(i, k)
	{
		for(int j = temp1; j < n; j++)
		{
			if (str[j] == 'o')
			{
				hataraku_L[j] = i + 1;
				temp1 = j + c + 1;
				ans[j][0] = true;
				break;
			}
		}
		for (int j = temp2; j >= 0; j--)
		{
			if (str[j] == 'o')
			{
				hataraku_R[j] = i + 1;
				temp2 = j - c - 1;
				ans[j][1] = true;
				break;
			}
		}
	}
	rep(i, n)
	{
		
		if (ans[i][0] && ans[i][1])
		{
			cout << i + 1 << endl;
		}
	}
	return 0;
}