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>
#define _LIBCPP_DEBUG 0
using namespace std;
typedef long long int ll;
ll MOD = 1e9 + 7;
int INF = 1 << 30;


int main(void){

    int N,K,C; cin>>N>>K>>C;
    string S; cin>>S;
    vector<int> SMAP(N,0);
	vector<int>SMAP2(N,0);
    int cnt=K;
	for (int i = N-1; i >= 0; i--) {
        if(S[i]=='o'){
            SMAP[i] = cnt;
            i -= C;
            cnt--;
        }
        if(cnt<=0) break;
    }
	/*
	for(int i = 0; i < N; i++){
		cout<<SMAP[i]<<endl;
	}
	cout<<"\n"<<endl;
	*/
	cnt = 1;
    for (int i = 0; i < N; ++i) {
        if(S[i]=='o'){
        	if(SMAP[i] == cnt) cout<<i+1<<endl;
        	SMAP2[i] = cnt;
            i += C;
            cnt++;
        }
        if(cnt>=K) break;
    }
	/*
	for(int i = 0; i < N; i++){
		cout<<SMAP2[i]<<endl;
	}
	*/
    /*
    for (int i = 0; i < N; ++i) {
        if(SMAP[i]==2) cout<<i+1<<endl;
    }
    cout<<endl;
	*/

    return 0;
}