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 ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define REP(i,j) for(int i=0;i<j;i++)
#define FORN(i,j,k) for(int i=j;i<k;i++)
#define vi vector<int>
#define vvi vector<vi >
#define pii pair<int,int>
#define vpii vector<pii >
using namespace std;
bool early[200005]={0},late[200005]={0};
int main(void){
	int n,k,c;
	string s;
	cin>>n>>k>>c;
	cin>>s;
	int last=-1,cnt=0;
	REP(i,n){
		if(s[i]=='o'){
			if(last==-1||i-last-1>=c){
				early[i]=1;
				last=i;
//				cout<<i<<' ';
				cnt++;
			}
		}
		if(cnt>=k)break;
	}
//	cout<<endl;
	last=-1;
	cnt=0;
	REP(i,n){
		if(s[n-i-1]=='o'){
			if(last==-1||last-(n-i-1)-1>=c){
				late[n-i-1]=1;
				last=n-i-1;
//				cout<<n-i-1<<' ';
				cnt++;
			}
		}
		if(cnt>=k)break;
	}
//	cout<<endl;
	REP(i,n){
		if(early[i]&&late[i]){
			cout<<i+1<<endl;
		}
	}
	return 0;
}