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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vvl = vector<vl>;

void ABC161E()
{
	int N, K, C;
	cin >> N >> K >> C;
	string S;
	cin >> S;
	vi X(N), Y(N);
	X[0] = S[0] == 'x' ? 0 : 1;
	X[1] = S[1] == 'x' ? 0 : 1;
	X[2] = S[2] == 'x' ? 0 : 1;
	X[3] = S[3] == 'x' ? 0 : 1;
	Y[N - 1] = S[N - 1] == 'x' ? 0 : 1;
	Y[N - 2] = S[N - 2] == 'x' ? 0 : 1;
	Y[N - 3] = S[N - 3] == 'x' ? 0 : 1;
	Y[N - 4] = S[N - 4] == 'x' ? 0 : 1;
	for (int i = 1; i < 4; ++i)
	{
		X[i] = max(X[i - 1], X[i]);
		Y[N - i - 1] = max(Y[N - i], Y[N - i - 1]);
	}
	for (int i = 4; i < N; ++i)
	{
		X[i] = max(X[i-1],X[i - 4] + (S[i] == 'x' ? 0 : 1));
		Y[N - i - 1] = max(Y[N-i],Y[N - i + 3] + (S[N - i - 1] == 'x' ? 0 : 1));
	}
	if (X[N - 1] != K)
		return;

	//for (int i = 0; i < 4; ++i)
	//{
	//	if(X[i] == 1 && Y[i] == K-1)
	//}

	for (int i = 0; i < N; ++i)
	{
		if (S[i] == 'x')
			continue;

		if((i == 0 || X[i] > X[i-1]) && (i == N-1 || Y[i] > Y[i+1]))
			cout << i + 1 << endl;
	}

}

int main()
{
	ABC161E();
	return 0;
}