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 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 400020
inline int read(){
    int x=0,f=1;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')f=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        x=(x<<3)+(x<<1)+c-'0';
        c=getchar();
    }
    return x*f;
}
char s[N];
int n,k,c,pos[N],dp[N],las[N],tot[N];
int main(){
	n=read(),k=read(),c=read();
	scanf("%s",s+1);
	int now=1;
	pos[0]=-c;
	int i=1;
	while(now<=n){
		while(now<=n&&s[now]=='x')las[now]=pos[i-1],tot[now]=i-1,++now;
		las[now]=pos[i-1],tot[now]=i-1;
		pos[i]=now;
		for(int j=1;j<=c+1;++j){
			++now;
			tot[now]=i,las[now]=pos[i];
		}
		++i;
	}
	for(int i=n;i>=1;--i){
		if(s[i]=='x')dp[i]=dp[i+1];
		else dp[i]=dp[i+c+1]+1;
	}
	for(int i=1;i<=n;++i){
		if(s[i]=='x')continue;
		if(tot[i]+dp[max(i+1,las[i]+c+1)]<k){
			printf("%d\n",i);
		}
	}
//	getchar();getchar();getchar();
	return 0;
}