#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(n);++i)
#define Rep(i,n) for(int i=1;i<=(n);++i)
using LL = long long;
using ULL = unsigned long long;

char s[200123];
int x[200123], y[200123];

int main()
{
	int n, k, c;
	cin >> n >> k >> c >> s+1;
	int cur = 1, cnt = 0;
	while(cnt < k) {
		if(s[cur] == 'o') x[++cnt] = cur, cur += c;
		++cur;
	}
	cur = n, cnt = k;
	while(cnt) {
		if(s[cur] == 'o') y[cnt--] = cur, cur -= c;
		--cur;
	}
	Rep(i, k) if(x[i] == y[i]) cout << x[i] << '\n';
	return 0;
}


### Prompt

Construct a Cpp code solution to the problem outlined:

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;
#define rep(i,n) for(int i=0;i<(n);++i)
#define Rep(i,n) for(int i=1;i<=(n);++i)
using LL = long long;
using ULL = unsigned long long;

char s[200123];
int x[200123], y[200123];

int main()
{
	int n, k, c;
	cin >> n >> k >> c >> s+1;
	int cur = 1, cnt = 0;
	while(cnt < k) {
		if(s[cur] == 'o') x[++cnt] = cur, cur += c;
		++cur;
	}
	cur = n, cnt = k;
	while(cnt) {
		if(s[cur] == 'o') y[cnt--] = cur, cur -= c;
		--cur;
	}
	Rep(i, k) if(x[i] == y[i]) cout << x[i] << '\n';
	return 0;
}


```