#include<bits/stdc++.h>
using namespace std;
#include<cstdio>
#define el "\n"
#define sp " "
typedef long long ll;

int main()
{
	ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n,k,c;
cin>>n>>k>>c;
string s;
cin>>s;
vector<ll> a(n),b(n);
ll cnt=1;
for(ll i=0;i<n;i++)
{
	if(s[i]=='o')
	{
		a[i]=cnt;
		cnt++;
		i+=c;
	}
	if(cnt>k)
	break;
}
cnt=k;
for(ll i=n-1;i>=0;i--)
{
	if(s[i]=='o')
	{
		b[i]=cnt;
		cnt--;
		i-=c;
	}
	if(cnt<1)
	break;
}
for(ll i=0;i<n;i++)
{
	if(a[i] && a[i]==b[i])
	cout<<i+1<<el;
}

}


### 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;
#include<cstdio>
#define el "\n"
#define sp " "
typedef long long ll;

int main()
{
	ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n,k,c;
cin>>n>>k>>c;
string s;
cin>>s;
vector<ll> a(n),b(n);
ll cnt=1;
for(ll i=0;i<n;i++)
{
	if(s[i]=='o')
	{
		a[i]=cnt;
		cnt++;
		i+=c;
	}
	if(cnt>k)
	break;
}
cnt=k;
for(ll i=n-1;i>=0;i--)
{
	if(s[i]=='o')
	{
		b[i]=cnt;
		cnt--;
		i-=c;
	}
	if(cnt<1)
	break;
}
for(ll i=0;i<n;i++)
{
	if(a[i] && a[i]==b[i])
	cout<<i+1<<el;
}

}


```