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)

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <math.h>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <functional>
#include <numeric>
#include <iostream>
#include <string>
#include <time.h>
#include <bitset>

using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }

int N, K, C;
string S;

int main() {

	cin >> N >> K >> C;
	cin >> S;

	vector<int> can1;
	vector<bool> worked(N);
	rep(i, N) {
		if (S[i] == 'o') {
			can1.push_back(i);
			worked[i] = true;
			i += C;
		}
	}

	if (can1.size() > K) {
		return 0;
	}

	vector<int> counter(N), maxim(N);
	rep(i, N) {
		if (i - C - 1 < 0) {
			if (S[i] == 'o') {
				counter[i] = 1;
			}
			else {
				counter[i] = 0;
			}
		}
		else {
			if (S[i] == 'o') {
				counter[i] = maxim[i - C - 1] + 1;
			}
			else {
				counter[i] = maxim[i - C - 1];
			}
		}
		if (i)
			maxim[i] = max(maxim[i - 1], counter[i]);
		else
			maxim[i] = counter[i];
	}

	vector<int> can2;
	rep(i, N) {
		int ri = N - 1 - i;

		if (S[ri] == 'o') {
			can2.push_back(ri);
			i += C;
		}
	}

	sort(all(can2));

	rep(i, K) if (can1[i] == can2[i]) cout << can1[i] + 1 << endl;



	return 0;
}