Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
	int N, K, C;
	char S[200001];
	scanf("%d %d %d", &N, &K, &C);
	scanf("%s", S);
	
	int i, k, dp[200002] = {};
	for (i = N, k = N * 2, dp[N+1] = 0; i >= 1; i--) {
		if (S[i-1] == 'o' && i + C < k) {
			dp[i] = dp[i+1] + 1;
			k = i;
		} else dp[i] = dp[i+1];
	}
	if (dp[1] > K) return 0;

	for (i = 1; i <= N; i++) {
		if (dp[i] > dp[i+1]) {
			printf("%d\n", i);
			i += C;
		} else if (S[i-1] == 'o') {
			for (k = i; k <= N && dp[k+1] == dp[i]; k++);
			i = (k > i + C)? k: i + C;
		} 
	}
	fflush(stdout);
	return 0;
}