// E - Yutori
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)

int main(){
	int N,K,C; string S;
	cin>>N>>K>>C>>S;

	vector<int> L,R;
	int c;

	c = 0;
	rep(i, N){
		if(c-- > 0) continue;
		c = 0;
		if(S[i] == 'o'){ L.push_back(i); c = C; }
	}

	c = 0;
	for(int i=N-1; i>=0; --i){
		if(c-- > 0) continue;
		c = 0;
		if(S[i] == 'o'){ R.push_back(i); c = C; }
	}

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

	if(L.size() > K) return 0; // 働ける日に余裕があると、その日に必ず働く必要がないから

	rep(i, L.size()) if(L[i] == R[i]) cout<<L[i]+1<<endl;
}
### Prompt

Generate a cpp solution to 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
// E - Yutori
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)

int main(){
	int N,K,C; string S;
	cin>>N>>K>>C>>S;

	vector<int> L,R;
	int c;

	c = 0;
	rep(i, N){
		if(c-- > 0) continue;
		c = 0;
		if(S[i] == 'o'){ L.push_back(i); c = C; }
	}

	c = 0;
	for(int i=N-1; i>=0; --i){
		if(c-- > 0) continue;
		c = 0;
		if(S[i] == 'o'){ R.push_back(i); c = C; }
	}

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

	if(L.size() > K) return 0; // 働ける日に余裕があると、その日に必ず働く必要がないから

	rep(i, L.size()) if(L[i] == R[i]) cout<<L[i]+1<<endl;
}
```