#include<bits/stdc++.h>
using namespace std;
int n,k,c;
char s[200005];
int l[200005],r[200005];
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>k>>c;
	for(int i=1; i<=n; i++)
		cin>>s[i];
	int cnt=1,nxt=1;
	for(int i=1; i<=n; i++) {
		if(s[i]=='o'&&i>=nxt) {
			l[cnt++]=i;
			nxt=i+c+1;
		}
	}
	cnt=k,nxt=n;
	for(int i=n; i>=1; i--) {
		if(s[i]=='o'&&i<=nxt) {
			r[cnt--]=i;
			nxt=i-c-1;
		}
		if(!cnt) break;
	}
	for(int i=1; i<=k; i++)
		if(l[i]==r[i])
			cout<<l[i]<<endl;
	return 0;
}

### Prompt

Your challenge is to write 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
#include<bits/stdc++.h>
using namespace std;
int n,k,c;
char s[200005];
int l[200005],r[200005];
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>k>>c;
	for(int i=1; i<=n; i++)
		cin>>s[i];
	int cnt=1,nxt=1;
	for(int i=1; i<=n; i++) {
		if(s[i]=='o'&&i>=nxt) {
			l[cnt++]=i;
			nxt=i+c+1;
		}
	}
	cnt=k,nxt=n;
	for(int i=n; i>=1; i--) {
		if(s[i]=='o'&&i<=nxt) {
			r[cnt--]=i;
			nxt=i-c-1;
		}
		if(!cnt) break;
	}
	for(int i=1; i<=k; i++)
		if(l[i]==r[i])
			cout<<l[i]<<endl;
	return 0;
}

```