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)

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define m_k make_pair
using namespace std;
const int N=2*1e5+100;
int n,k,c,dp[N*2],a[N],w;
char s[N];
signed main()
{
	scanf("%d%d%d",&n,&k,&c);
	scanf("%s",s+1);
	dp[n+1]=0;
	if (s[n]=='o') dp[n]=1;
	else dp[n]=0;
	for (int i=n-1;i>=1;i--)
	{
		dp[i]=dp[i+1];
		if (s[i]=='o')
		  dp[i]=max(dp[i],dp[i+c+1]+1);
	}
	for (int i=1;i<=n;i++)
	{
		if (s[i]=='o')
		{
			w++;
			a[w]=i;
			if (w==k) break;
			i=i+c;
		}
	}
//	for (int i=1;i<=n;i++) printf("%d ",dp[i]);
//	printf("\n");
	for (int i=1;i<=n;i++)
	{
		if (s[i]=='o')
		{
			if (i>a[w]) continue;
			int pos=lower_bound(a+1,a+1+w,i)-a;
			if (a[pos]!=i) continue;
			if (pos-1+dp[i+1]<k) printf("%d\n",i);
		}
	}
}