#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
typedef long long int LL;

int n, k, c;
string s;
int l[maxn], r[maxn];

int main(){
	scanf("%d %d %d", &n, &k, &c);
    cin>>s;
    
	memset(l, 0, sizeof(l));
	memset(r, 0, sizeof(r));
	int cnt = 0;
	for(int i = 0; i < n; i++){
		if(s[i] == 'o'){
			l[i] = ++cnt;
			i += c;
		}
	}
	cnt = k;
	for(int i = n-1; i >= 0; i--){
		if(s[i] == 'o'){
			r[i] = cnt--;
			i -= c;
		}
	}
	for(int i = 0; i <n; i++){
		if(r[i] == l[i] && r[i] > 0){
			printf("%d\n", 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 = 1e6 + 5;
typedef long long int LL;

int n, k, c;
string s;
int l[maxn], r[maxn];

int main(){
	scanf("%d %d %d", &n, &k, &c);
    cin>>s;
    
	memset(l, 0, sizeof(l));
	memset(r, 0, sizeof(r));
	int cnt = 0;
	for(int i = 0; i < n; i++){
		if(s[i] == 'o'){
			l[i] = ++cnt;
			i += c;
		}
	}
	cnt = k;
	for(int i = n-1; i >= 0; i--){
		if(s[i] == 'o'){
			r[i] = cnt--;
			i -= c;
		}
	}
	for(int i = 0; i <n; i++){
		if(r[i] == l[i] && r[i] > 0){
			printf("%d\n", i+1);
		}
	}
}

```