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 <stdio.h>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long LL;

const int maxn=100000+10;
int L[maxn][2],R[maxn][2];
char str[maxn];

int main()
{
    int i,j,n,K,C;
    scanf("%d%d%d",&n,&K,&C);
    scanf("%s",str+1);
    for(i=1;i<=n;i++){
        if(str[i]=='x'){
            L[i][0]=max(L[i-1][0],L[i-1][1]);
            L[i][1]=L[i][0];
        }else {
            L[i][0]=max(L[i-1][0],L[i-1][1]);
            int v=0;
            if(i-C>1)
                v=L[i-C-1][1];
            L[i][1]=v+1;
        }
        L[i][0]=max(L[i][0],L[i-1][0]);
        L[i][1]=max(L[i][1],L[i-1][1]);
        //printf("i=%d L0=%d L1=%d\n",i,L[i][0],L[i][1]);
    }
    for(i=n;i>=1;i--){
        if(str[i]=='x'){
            R[i][0]=max(R[i+1][0],R[i+1][1]);
            R[i][1]=R[i][0];
        }else {
            R[i][0]=max(R[i+1][0],R[i+1][1]);
            int v=0;
            if(i+C+1<=n)
                v=max(R[i+C+1][1],R[i+C+1][1]);
            R[i][1]=v+1;
        }
        R[i][0]=max(R[i][0],R[i+1][0]);
        R[i][1]=max(R[i][1],R[i+1][1]);
        //printf("i=%d R0=%d R1=%d\n",i,R[i][0],R[i][1]);
    }
    
    for(i=1;i<=n;i++){
        if(str[i]=='x')
            continue;
        if(L[i][0]+R[i][0]==K-1)
            printf("%d\n",i);
    }
    return 0;
}