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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
	problem: 
	
	
*/

#include<bits/stdc++.h>
/*
#pragma GCC optamize("O2")
#pragma G++ optamize("O2")
*/
using namespace std;
const long long maxn = 0x3f3f3f3f;
typedef long long LL;

LL N, K, C;
string s;
LL ans[200100];

int main() {
	
//	freopen(".in", "r", stdin);
//	freopen(".out", "w", stdout);
	
	LL i, j, k;
    scanf("%lld%lld%lld", &N, &K, &C);
    cin >> s;
    LL cnt = 0;
    for (i = 0; i < N; ++ i) {
        if (s[i] == 'o' && cnt < K) {
            ++ ans[i];
            ++ cnt;
            i = i + C;
        }
        if (cnt >= K) break;
    }
    LL cnt1 = 0;
    for (i = N - 1; i >= 0; -- i) {
        if (s[i] == 'o' && cnt1 < K) {
            ++ cnt1;
            ++ ans[i];
            i = i - C;
        }
        if (cnt1 >= K) break;
    }
    for (i = 0; i < N; ++ i) {
        if (ans[i] > 1) printf("%lld\n", i + 1);
    }
	
	return 0;
}
/*
11 3 2
ooxxxoxxxoo
5 2 3
ooxoo
*/