#include <bits/stdc++.h>
#define ll long long

using namespace std;

int main()
{

	int n, k, c;
	cin >> n >> k >> c;
	string s;
	cin >> s;
	vector<int> A(n,0);
	vector<int> B(n,0);
	vector<int> C;
	int  a=0;
	for(int i=0; i<n; i++){
		if(s[i]=='o'){
			A[i]++;
			i=i+c;
			a++;
		}
	}

	for(int i=(n-1); i>=0; i--){
		if(s[i]=='o'){
			B[i]++;
			i=i-c;
		}
	}
	
	
	for(int i=0; i<A.size(); i++){
		if(A[i]==B[i] && A[i]>0){
			C.push_back(i+1);
		}
	}
		
	if(a>(c+k)){
		cout<<" ";
	}
	else{
		for(int i=0; i<C.size(); i++){
			cout<<C[i]<<endl;
		}
	}

}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
#define ll long long

using namespace std;

int main()
{

	int n, k, c;
	cin >> n >> k >> c;
	string s;
	cin >> s;
	vector<int> A(n,0);
	vector<int> B(n,0);
	vector<int> C;
	int  a=0;
	for(int i=0; i<n; i++){
		if(s[i]=='o'){
			A[i]++;
			i=i+c;
			a++;
		}
	}

	for(int i=(n-1); i>=0; i--){
		if(s[i]=='o'){
			B[i]++;
			i=i-c;
		}
	}
	
	
	for(int i=0; i<A.size(); i++){
		if(A[i]==B[i] && A[i]>0){
			C.push_back(i+1);
		}
	}
		
	if(a>(c+k)){
		cout<<" ";
	}
	else{
		for(int i=0; i<C.size(); i++){
			cout<<C[i]<<endl;
		}
	}

}

```