#include <bits/stdc++.h>

using namespace std;

int n,k,c;
char s[200005];
int f[500005];

int main()
{
	scanf("%d%d%d",&n,&k,&c);
	scanf("%s",s+1);
	memset(f,0,sizeof(f));
	for (int i=n;i>=1;--i)
	{
		f[i]=f[i+1];
		if (s[i]=='x') continue;
		f[i]=max(f[i],1+f[i+c+1]);
	}
	int now=0,last=-1e9;
	vector<int> v;
	v.clear();
	for (int i=1;i<=n;++i)
	{
		if (s[i]=='x') continue;
		if (f[i+1]+now>=k)
		{
			if (i-last>=c+1) now++,last=i;
		} else
		{
			v.push_back(i);
			now++;
			last=i;
		}
	}
	for (int i=0;i<v.size();++i) printf("%d\n",v[i]);
	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;

int n,k,c;
char s[200005];
int f[500005];

int main()
{
	scanf("%d%d%d",&n,&k,&c);
	scanf("%s",s+1);
	memset(f,0,sizeof(f));
	for (int i=n;i>=1;--i)
	{
		f[i]=f[i+1];
		if (s[i]=='x') continue;
		f[i]=max(f[i],1+f[i+c+1]);
	}
	int now=0,last=-1e9;
	vector<int> v;
	v.clear();
	for (int i=1;i<=n;++i)
	{
		if (s[i]=='x') continue;
		if (f[i+1]+now>=k)
		{
			if (i-last>=c+1) now++,last=i;
		} else
		{
			v.push_back(i);
			now++;
			last=i;
		}
	}
	for (int i=0;i<v.size();++i) printf("%d\n",v[i]);
	return 0;
}
```