#include <bits/stdc++.h>
using namespace std;

int main() {
	int n, i, k, c, dp[222222];
	string s;
	vector<int> v;
	cin >> n >> k >> c >> s;
	dp[0] = 0;
	for (i = 1; i <= n; i++)
	{
		if(s[i-1]=='o'){
			if(i-c-1<0)
				dp[i] = 1;
			else
				dp[i] = dp[i - c - 1] + 1;
		}
		else
			dp[i] = dp[i - 1];
	}
	i = n;
	while(s[i-1]=='x')
		i--;
	while (k > 0)
	{
		int p = i;
		i--;
		while (s[i - 1] == 'x' && i > 1)
			i--;
		if(dp[i]<k)
			v.push_back(p);
		while((i+c>=p||s[i-1]=='x') && i > 1)
			i--;
		k--;
	}
	sort(v.begin(), v.end());
	for(auto &e:v)
		cout << e << endl;
}
### Prompt

Please provide a cpp coded solution 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;

int main() {
	int n, i, k, c, dp[222222];
	string s;
	vector<int> v;
	cin >> n >> k >> c >> s;
	dp[0] = 0;
	for (i = 1; i <= n; i++)
	{
		if(s[i-1]=='o'){
			if(i-c-1<0)
				dp[i] = 1;
			else
				dp[i] = dp[i - c - 1] + 1;
		}
		else
			dp[i] = dp[i - 1];
	}
	i = n;
	while(s[i-1]=='x')
		i--;
	while (k > 0)
	{
		int p = i;
		i--;
		while (s[i - 1] == 'x' && i > 1)
			i--;
		if(dp[i]<k)
			v.push_back(p);
		while((i+c>=p||s[i-1]=='x') && i > 1)
			i--;
		k--;
	}
	sort(v.begin(), v.end());
	for(auto &e:v)
		cout << e << endl;
}
```