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)

#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define ll long long
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB emplace_back
#define PF emplace_front
#define FILL(a,b) memset(a,b,sizeof(a))
#define lwb lower_bound
#define upb upper_bound
using namespace std;
int l[200005],r[200005],la[200005],ra[200005],n,k,c;
char s[200005];
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n>>k>>c;
	c++;
	FOR(i,1,n)cin>>s[i];
	FOR(i,1,n){
		if(s[i-1]=='o')l[i]=i-1;
		else l[i]=l[i-1];
		if(s[i]=='o'){
			if(i>c)la[i]=la[i-c]+1;
			else la[i]=1;
		}else la[i]=la[i-1];
	}
	for(int i=n;i>=1;i--){
		if(s[i+1]=='o')r[i]=i+1;
		else r[i]=r[i+1];
		if(s[i]=='o'){
			if(n-i>=c)ra[i]=ra[i+c]+1;
			else ra[i]=1;
		}else ra[i]=ra[i+1];
	}
	FOR(i,1,n){
		if(la[l[i]]+ra[r[i]]==k-1)cout<<i<<'\n';
	}
	RE 0;
}