#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int INF=1<<29;

int main(){
	int n,k,c;
	string s; cin>>n>>k>>c>>s;

	auto solve=[&](){
		vector<int> res;
		int pre=-INF;
		rep(i,n) if(s[i]=='o' && pre+c<i) {
			res.emplace_back(i);
			pre=i;
		}
		return res;
	};

	vector<int> L=solve();
	if(L.size()>k) return 0;

	reverse(s.begin(),s.end());

	vector<int> R=solve();
	rep(i,R.size()) R[i]=n-R[i]-1;
	reverse(R.begin(),R.end());

	rep(i,L.size()) if(L[i]==R[i]) printf("%d\n",L[i]+1);

	return 0;
}

### 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 rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

const int INF=1<<29;

int main(){
	int n,k,c;
	string s; cin>>n>>k>>c>>s;

	auto solve=[&](){
		vector<int> res;
		int pre=-INF;
		rep(i,n) if(s[i]=='o' && pre+c<i) {
			res.emplace_back(i);
			pre=i;
		}
		return res;
	};

	vector<int> L=solve();
	if(L.size()>k) return 0;

	reverse(s.begin(),s.end());

	vector<int> R=solve();
	rep(i,R.size()) R[i]=n-R[i]-1;
	reverse(R.begin(),R.end());

	rep(i,L.size()) if(L[i]==R[i]) printf("%d\n",L[i]+1);

	return 0;
}

```