#include <bits/stdc++.h>
using namespace std;
 
const int N=200005;

int n,k,c;
int a[N],b[N];
char s[N];

void init()
{scanf("%d %d %d",&n,&k,&c);
 scanf("%s",s+1);
}

void work()
{int pos=0;
 for (int i=1;i<=n;i++)
   if (s[i]=='o')
     {a[++pos]=i;
      if (pos>=k) break;
      i+=c;
     }
 pos=k;
 for (int i=n;i>=1;i--)
   if (s[i]=='o')
     {b[pos--]=i;
      if (pos<=0) break;
      i-=c;
	 }
 for (int i=1;i<=k;i++)
   if (a[i]==b[i])
     printf("%d\n",a[i]);  
}

int main()
{
 init();
 work();
 return 0;
}

### 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
#include <bits/stdc++.h>
using namespace std;
 
const int N=200005;

int n,k,c;
int a[N],b[N];
char s[N];

void init()
{scanf("%d %d %d",&n,&k,&c);
 scanf("%s",s+1);
}

void work()
{int pos=0;
 for (int i=1;i<=n;i++)
   if (s[i]=='o')
     {a[++pos]=i;
      if (pos>=k) break;
      i+=c;
     }
 pos=k;
 for (int i=n;i>=1;i--)
   if (s[i]=='o')
     {b[pos--]=i;
      if (pos<=0) break;
      i-=c;
	 }
 for (int i=1;i<=k;i++)
   if (a[i]==b[i])
     printf("%d\n",a[i]);  
}

int main()
{
 init();
 work();
 return 0;
}

```