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

const int maxn = 200005;

int n, k, c, d, pos;
int l[maxn], r[maxn];
char s[maxn];

signed main () {
	scanf("%d %d %d", &n, &k, &c);
	scanf("%s", &s);
	d = 1, pos = 0;
	while (d <= k && pos < n) {
		while (s[pos] == 'x' && pos < n) 
			++pos;
		l[d] = pos;
		pos += c + 1, ++d;
	}
	d = k, pos = n - 1;
	while (d && pos >= 0) {
		while (s[pos] == 'x' && pos >= 0)
			--pos;
		r[d] = pos;
		pos -= c + 1, --d;
	}
	for (int i = 1; i <= k; ++i) {
		if (l[i] == r[i])
			printf("%d\n", l[i] + 1);
	}
}

### 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 maxn = 200005;

int n, k, c, d, pos;
int l[maxn], r[maxn];
char s[maxn];

signed main () {
	scanf("%d %d %d", &n, &k, &c);
	scanf("%s", &s);
	d = 1, pos = 0;
	while (d <= k && pos < n) {
		while (s[pos] == 'x' && pos < n) 
			++pos;
		l[d] = pos;
		pos += c + 1, ++d;
	}
	d = k, pos = n - 1;
	while (d && pos >= 0) {
		while (s[pos] == 'x' && pos >= 0)
			--pos;
		r[d] = pos;
		pos -= c + 1, --d;
	}
	for (int i = 1; i <= k; ++i) {
		if (l[i] == r[i])
			printf("%d\n", l[i] + 1);
	}
}

```