#include <bits/stdc++.h>
using namespace std;
#define pb push_back

int main(void)
{
	int n, k, c; cin >> n >> k >> c;
	string s; cin >> s;
	vector<int> left, right;
	for(int i=0;i<n;i+=(c+1)) {
		while(s[i]=='x') i++;
		right.pb(i);
		if(right.size() == k) break;
	}
	for(int i=n-1;i>=0;i-=(c+1)) {
		while(s[i]=='x') i--;
		left.pb(i);
		if(left.size() == k) break;
	}
	if(right.size() < k || left.size() < k) return 0;
	reverse(left.begin(), left.end());
	for(int i=0;i<k;i++) {
		if(left[i] == right[i]) {
			cout << left[i] + 1 << endl;
		}
	}


	return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 pb push_back

int main(void)
{
	int n, k, c; cin >> n >> k >> c;
	string s; cin >> s;
	vector<int> left, right;
	for(int i=0;i<n;i+=(c+1)) {
		while(s[i]=='x') i++;
		right.pb(i);
		if(right.size() == k) break;
	}
	for(int i=n-1;i>=0;i-=(c+1)) {
		while(s[i]=='x') i--;
		left.pb(i);
		if(left.size() == k) break;
	}
	if(right.size() < k || left.size() < k) return 0;
	reverse(left.begin(), left.end());
	for(int i=0;i<k;i++) {
		if(left[i] == right[i]) {
			cout << left[i] + 1 << endl;
		}
	}


	return 0;
}

```