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: Accepted)

#include <iostream>
#include <algorithm>
#include <string>
using namespace std;
typedef long long ll;


int main(void) {
	int n, k, c, i, g1[200003], g2[200003], d1[200003], d2[200003], rs;
	string s;
	bool ir1[200003] = { false }, ir2[200003] = { false };

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

	i = 0;
	while (i < n) {
		while (i < n && s[i] == 'x') i++;
		if (i == n) break;
		ir1[i] = true;
		i += c + 1;
	}

	i = n - 1;
	while (i >= 0) {
		while (i >= 0 && s[i] == 'x') i--;
		if (i == -1) break;
		ir2[i] = true;
		i -= c + 1;
	}

	if (ir1[0]) {
		d1[0] = 1;
	}
	for (i = 1; i < n; i++) {
		if (ir1[i]) {
			d1[i] = d1[i - 1] + 1;
			g1[i] = i;
		}
		else {
			d1[i] = d1[i - 1];
			g1[i] = g1[i - 1];
		}
	}

	if (ir2[n - 1]) {
		d2[n - 1] = 1;
		g2[n - 1] = n - 1;
	}
	for (i = n - 2; i >= 0; i--) {
		if (ir2[i]) {
			d2[i] = d2[i + 1] + 1;
			g2[i] = i;
		}
		else {
			d2[i] = d2[i + 1];
			g2[i] = g2[i + 1];
		}
	}

	for (i = 0; i < n; i++) {
		if (s[i] == 'x') continue;
		rs = 0;
		if (i > 0) rs += d1[i - 1];
		if (i < n - 1) rs += d2[i + 1];
		if (i > 0 && i < n - 1) {
			if (d1[i - 1] > 0 && d2[i + 1] > 0 && g2[i + 1] - g1[i - 1] <= c) {
				rs--;
			}
		}
		if (rs < k) cout << i + 1 << endl;
	}

	return 0;
}