#include <iostream>
#include <cstdio>

using namespace std;

const int N = 200005;

int n, K, C, f[N], g[N];

char s[N];

int main() {
	scanf("%d%d%d%s", &n, &K, &C, s + 1);
	int v = 0;
	for (int i = 1; i <= n; i++) {
		f[i] = f[i - 1];
		if (s[i] == 'x') continue;
		if (i - C - 1 >= 1) v = max(v, f[i - C - 1]);
		f[i] = max(f[i], v + 1);
	}
	v = 0;
	for (int i = n; i; i--) {
		g[i] = g[i + 1];
		if (s[i] == 'x') continue;
		if (i + C + 1 <= n) v = max(v, g[i + C + 1]);
		g[i] = max(g[i], v + 1);
	}


	for (int i = 1; i <= n; i++)
		if (f[i - 1] + g[i + 1] < K) printf("%d\n", i);
	return 0;
}

### 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 <iostream>
#include <cstdio>

using namespace std;

const int N = 200005;

int n, K, C, f[N], g[N];

char s[N];

int main() {
	scanf("%d%d%d%s", &n, &K, &C, s + 1);
	int v = 0;
	for (int i = 1; i <= n; i++) {
		f[i] = f[i - 1];
		if (s[i] == 'x') continue;
		if (i - C - 1 >= 1) v = max(v, f[i - C - 1]);
		f[i] = max(f[i], v + 1);
	}
	v = 0;
	for (int i = n; i; i--) {
		g[i] = g[i + 1];
		if (s[i] == 'x') continue;
		if (i + C + 1 <= n) v = max(v, g[i + C + 1]);
		g[i] = max(g[i], v + 1);
	}


	for (int i = 1; i <= n; i++)
		if (f[i - 1] + g[i + 1] < K) printf("%d\n", i);
	return 0;
}

```