#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int MOD = 1e9+7;
 
int main(){
	
	int n,k,c; string s;
	cin>>n>>k>>c>>s;
	vector<int> t1(n),t2(n);
	for(int i=0; i<n; i++){
		t1.at(i)=n*2+1;
		t2.at(i)=n*3+1;
	}
	int cnt{},i{};
	while(i<n){
		if(s.at(i)=='o'){
			cnt++;
			t1.at(i)=cnt;
			if(cnt==k) break;
			i+=c+1;
		}
		else i++;
	}
	i=n-1;
	while(i>=0){
		if(s.at(i)=='o'){
			t2.at(i)=cnt;
			cnt--;
			if(cnt==-1) break;
			i-=c+1;
		}
		else i--;
	}
	for(int i=0; i<n; i++){
		if(t1.at(i)==t2.at(i)) cout<<i+1 << endl;
	}
	
}

### Prompt

In Cpp, your task is to solve the following problem:

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.

Examples

Input

11 3 2
ooxxxoxxxoo


Output

6


Input

5 2 3
ooxoo


Output

1
5


Input

5 1 0
ooooo


Output




Input

16 4 3
ooxxoxoxxxoxoxxo


Output

11
16

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
const int MOD = 1e9+7;
 
int main(){
	
	int n,k,c; string s;
	cin>>n>>k>>c>>s;
	vector<int> t1(n),t2(n);
	for(int i=0; i<n; i++){
		t1.at(i)=n*2+1;
		t2.at(i)=n*3+1;
	}
	int cnt{},i{};
	while(i<n){
		if(s.at(i)=='o'){
			cnt++;
			t1.at(i)=cnt;
			if(cnt==k) break;
			i+=c+1;
		}
		else i++;
	}
	i=n-1;
	while(i>=0){
		if(s.at(i)=='o'){
			t2.at(i)=cnt;
			cnt--;
			if(cnt==-1) break;
			i-=c+1;
		}
		else i--;
	}
	for(int i=0; i<n; i++){
		if(t1.at(i)==t2.at(i)) cout<<i+1 << endl;
	}
	
}

```