#include <bits/stdc++.h>
using namespace std;
int main() {
	int N, K, C;
	string S;
	cin >> N >> K >> C >> S;
	vector<int> a(N), b(N);
	for (int i = 0, j = 0; i < N && j < K; ) {
		if (S[i] == 'o') {
			a[i] = ++ j;
			i += C+1;
		} else {
			++ i;
		}
	}
	for (int i = N-1, j = 0; i >= 0 && j < K; ) {
		if (S[i] == 'o') {
			b[i] = ++ j;
			i -= C+1;
		} else {
			-- i;
		}
	}
	for (int i = 0; i < N; ++ i) {
		// cerr << a[i] << " " << b[i] << endl;
		if (a[i] && b[i] && a[i]+b[i] == K+1) {
			cout << i+1 << endl;
		}
	}
}

### 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;
int main() {
	int N, K, C;
	string S;
	cin >> N >> K >> C >> S;
	vector<int> a(N), b(N);
	for (int i = 0, j = 0; i < N && j < K; ) {
		if (S[i] == 'o') {
			a[i] = ++ j;
			i += C+1;
		} else {
			++ i;
		}
	}
	for (int i = N-1, j = 0; i >= 0 && j < K; ) {
		if (S[i] == 'o') {
			b[i] = ++ j;
			i -= C+1;
		} else {
			-- i;
		}
	}
	for (int i = 0; i < N; ++ i) {
		// cerr << a[i] << " " << b[i] << endl;
		if (a[i] && b[i] && a[i]+b[i] == K+1) {
			cout << i+1 << endl;
		}
	}
}

```