#include <iostream>
using namespace std;

string s;
int a[222222], b[222222];

int main(void)
{
	int n, k, c, i;

	cin >> n >> k >> c >> s;
	s = '.' + s;

	int cnta = 0;
	a[0] = -1e8;
	for (i = 1; i <= n; i++)
		if (s[i] == 'o')
			if (i - a[cnta] > c)
				a[++cnta] = i;

	int cntb = 0;
	b[0] = 1e8;
	for (i = n; i >= 1; i--)
		if (s[i] == 'o')
			if (b[cntb] - i > c)
				b[++cntb] = i;

	for (i = 1; i <= min(cnta, k); i++)
		if (a[i] == b[min(cntb, k) - i + 1])
			cout << a[i] << "\n";
	return 0;
}
### Prompt

In CPP, your task is to solve 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 <iostream>
using namespace std;

string s;
int a[222222], b[222222];

int main(void)
{
	int n, k, c, i;

	cin >> n >> k >> c >> s;
	s = '.' + s;

	int cnta = 0;
	a[0] = -1e8;
	for (i = 1; i <= n; i++)
		if (s[i] == 'o')
			if (i - a[cnta] > c)
				a[++cnta] = i;

	int cntb = 0;
	b[0] = 1e8;
	for (i = n; i >= 1; i--)
		if (s[i] == 'o')
			if (b[cntb] - i > c)
				b[++cntb] = i;

	for (i = 1; i <= min(cnta, k); i++)
		if (a[i] == b[min(cntb, k) - i + 1])
			cout << a[i] << "\n";
	return 0;
}
```