#include <bits/stdc++.h>

#define MAXN 200010
#define ll long long
#define mod 1000000007

using namespace std;

int n, m, k, c;
int ans, tot;

int a[MAXN];

string s;

int main(){
	int t;
	scanf("%d%d%d", &n, &k, &c);
	cin >> s;
	
	
	int i = n, num = 0;
	while (i >= 1) {
		if (s[i - 1] == 'o') {
			num++;
			a[i]++;
		i -= c;
		}
		i--;
	}
	
//	for (int i = 1; i <= n; i++) {
//		cout << a[i] << " ";
//	}
//	cout << endl;
	
	if (num != k) {
//		cout << num << endl;
		return 0;
	}
	
	i = 1;
	while (i <= n) {
		if (s[i - 1] == 'o') {
			if (a[i] == 1) cout << i << endl;
		i += c;
		}
		i++;
	}

	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 <bits/stdc++.h>

#define MAXN 200010
#define ll long long
#define mod 1000000007

using namespace std;

int n, m, k, c;
int ans, tot;

int a[MAXN];

string s;

int main(){
	int t;
	scanf("%d%d%d", &n, &k, &c);
	cin >> s;
	
	
	int i = n, num = 0;
	while (i >= 1) {
		if (s[i - 1] == 'o') {
			num++;
			a[i]++;
		i -= c;
		}
		i--;
	}
	
//	for (int i = 1; i <= n; i++) {
//		cout << a[i] << " ";
//	}
//	cout << endl;
	
	if (num != k) {
//		cout << num << endl;
		return 0;
	}
	
	i = 1;
	while (i <= n) {
		if (s[i - 1] == 'o') {
			if (a[i] == 1) cout << i << endl;
		i += c;
		}
		i++;
	}

	return 0;
}

```