#include<iostream>
#include<vector>
using namespace std;
int main() {
	int n, k, c; string s;
	cin >> n >> k >> c >> s;
	int* l = new int[k];
	int* r = new int[k];
	for (int i = 0,ki=0; i < n; i++) {
		if (s[i] == 'o') {
			l[ki++] = i; i += c;
			if (ki == k) break;
		}
	}
	for (int i = n-1,ki=k-1; i >= 0; i--) {
		if (s[i] == 'o') {
			r[ki--] = i; i -= c;
			if (ki == -1) break;
		}
	}
	for (int x = 0; x < k; x++) {
		if (l[x] == r[x]) cout << l[x] + 1 << endl;
	}
	return 0;
}

### Prompt

Please formulate 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<iostream>
#include<vector>
using namespace std;
int main() {
	int n, k, c; string s;
	cin >> n >> k >> c >> s;
	int* l = new int[k];
	int* r = new int[k];
	for (int i = 0,ki=0; i < n; i++) {
		if (s[i] == 'o') {
			l[ki++] = i; i += c;
			if (ki == k) break;
		}
	}
	for (int i = n-1,ki=k-1; i >= 0; i--) {
		if (s[i] == 'o') {
			r[ki--] = i; i -= c;
			if (ki == -1) break;
		}
	}
	for (int x = 0; x < k; x++) {
		if (l[x] == r[x]) cout << l[x] + 1 << endl;
	}
	return 0;
}

```