#include <bits/stdc++.h>
using namespace std;
const int MAXN=200010;
int n,k,c,tot,l[MAXN],r[MAXN];
char s[MAXN];
int main () {
	scanf("%d%d%d%s",&n,&k,&c,s+1);
	int las=-c;
	for (int i=1;i<=n;i++) {
		if (s[i]=='x'||i-las<c+1) {continue;}
		l[++tot]=las=i;
	}
	if (tot!=k) {return 0;}
	tot=k+1;
	las=n+c+1;
	for (int i=n;i>=1;i--) {
		if (s[i]=='x'||las-i<c+1) {continue;}
		r[--tot]=las=i;
	}
	for (int i=1;i<=k;i++) {
		if (l[i]==r[i]) {printf("%d\n",l[i]);}
	}
	return 0;
}
### Prompt

Develop a solution in cpp to the problem described below:

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=200010;
int n,k,c,tot,l[MAXN],r[MAXN];
char s[MAXN];
int main () {
	scanf("%d%d%d%s",&n,&k,&c,s+1);
	int las=-c;
	for (int i=1;i<=n;i++) {
		if (s[i]=='x'||i-las<c+1) {continue;}
		l[++tot]=las=i;
	}
	if (tot!=k) {return 0;}
	tot=k+1;
	las=n+c+1;
	for (int i=n;i>=1;i--) {
		if (s[i]=='x'||las-i<c+1) {continue;}
		r[--tot]=las=i;
	}
	for (int i=1;i<=k;i++) {
		if (l[i]==r[i]) {printf("%d\n",l[i]);}
	}
	return 0;
}
```