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 5.4.1) (judge verdict: Accepted)

int N,K,C;
char*S;
int R[2<<17];
char buf[2<<17],out[3<<19];sz;outsz;
main()
{
	read(0,buf,2<<17);
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		N=N*10+k-'0';
	}
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		K=K*10+k-'0';
	}
	for(;;)
	{
		char k=buf[sz++];
		if(k<'0')break;
		C=C*10+k-'0';
	}
	S=buf+sz;
	int now=K;
	for(int i=N;i-->0;)
	{
		if(S[i]=='o')
		{
			R[i]=now--;
			i-=C;
		}
	}
	now=0;
	for(int i=0;i<N;i++)
	{
		if(S[i]=='o')
		{
			if(R[i]==++now)
			{
				int t=i+1;
				if(t<10)out[outsz++]=t+'0';
				else if(t<100)
				{
					out[outsz++]=t/10+'0';
					out[outsz++]=t%10+'0';
				}
				else if(t<1000)
				{
					out[outsz++]=t/100+'0';
					out[outsz++]=t/10%10+'0';
					out[outsz++]=t%10+'0';
				}
				else if(t<10000)
				{
					out[outsz++]=t/1000+'0';
					out[outsz++]=t/100%10+'0';
					out[outsz++]=t/10%10+'0';
					out[outsz++]=t%10+'0';
				}
				else if(t<100000)
				{
					out[outsz++]=t/10000+'0';
					out[outsz++]=t/1000%10+'0';
					out[outsz++]=t/100%10+'0';
					out[outsz++]=t/10%10+'0';
					out[outsz++]=t%10+'0';
				}
				else
				{
					out[outsz++]=t/100000+'0';
					out[outsz++]=t/10000%10+'0';
					out[outsz++]=t/1000%10+'0';
					out[outsz++]=t/100%10+'0';
					out[outsz++]=t/10%10+'0';
					out[outsz++]=t%10+'0';
				}
				out[outsz++]='\n';
			}
			i+=C;
		}
	}
	write(1,out,outsz);
}