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 ll long long
#define ull unsigned long long
#define MP make_pair
#define PII pair<int, int>
#define PLL pair<ll, ll>
#define PLI pair<ll, int>
#define PIL pair<int, ll>
#define STR string
#define Fst first
#define Snd second
#define SHN 1e9
#define LSHN 1e18
using namespace std;
int n,m,k;
string s;
int mx[800010];
vector<int> v[200010];
int l[200010],r[200010];
int siz=1;
int search(int x,int y)
{
	if (x>y) return 0;
	x+=siz/2,y+=siz/2+1;
	int res=0;
	while (x<y)
	{
		if (x%2) res=max(res,mx[x++]); 
		if (y%2) res=max(res,mx[--y]); 
		x/=2,y/=2;
	} 
	return res;
}
void update(int x,int y)
{
	x+=siz/2;
	while (x) mx[x]=max(mx[x],y),x/=2;
}
int main()
{
	cin >>n>>m>>k>>s;
	while (siz<2*(n+2)) siz*=2;
	for (int i=0;i<n;i++)
	{
		l[i]=search(0,i-k-1)+(s[i]=='o');
		update(i,l[i]);
	}
	memset(mx,0,sizeof(mx));
	for (int i=n-1;i>=0;i--)
	{
		r[i]=search(i+k+1,n-1)+(s[i]=='o');
		update(i,r[i]);
	}
	for (int i=1;i<n;i++) l[i]=max(l[i],l[i-1]); 
	for (int i=n-2;i>=0;i--) r[i]=max(r[i],r[i+1]); 
	for (int i=0;i<n;i++)
	{
		int li=(i-k-1<0 ? 0 : l[i-k-1]);
		int ri=(i+k+1>=n ? 0 : r[i+k+1]);
		//cout <<i<<" "<<li<<" "<<ri<<" ##\n";
		if (s[i]=='o' && li+ri==m-1) v[li].push_back(i+1);
	}
	for (int i=0;i<=m;i++)
	{
		//cout <<i<<" "<<v[i].size()<<" ##\n"; 
		if (v[i].size()==1) cout <<v[i][0]<<"\n";
	}
	return 0; 
}