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: Wrong Answer)

#include <cstdio>
#include <iostream>
#include <cmath>
#include <set>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
const int N=400000+10;
int dp[N][2],n,k,c;
char s[N];
bool vis[N][2];
int main()
{
    //freopen("a.in","r",stdin);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k>>c;
    c++;
    scanf("%s",s+1);
    for(int i=1;i<=n;i++){
    	if(s[i]=='o'){
    		dp[i][0]=i-c>=1?dp[i-c][0]+1:1;
    		if(dp[i][0]>dp[i-1][0])vis[i][0]=true;
    		dp[i][0]=max(dp[i-1][0],dp[i][0]);
    	}
    	else dp[i][0]=max(dp[i-1][0],dp[i][0]);
    }
    for(int i=n;i>=1;i--){
    	if(s[i]=='o'){
    		dp[i][1]=dp[i+c][1]+1;
    		if(dp[i][1]>dp[i+1][1])vis[i][1]=true;
    		dp[i][1]=max(dp[i][1],dp[i+1][1]);
    	}
    	else dp[i][1]=max(dp[i][1],dp[i+1][1]);
    }
    vector<int> ans;
    for(int i=1;i<=n;i++){
    	if(dp[i][0]+dp[i][1]==k+1&&s[i]=='o'&&(vis[i][0]&&vis[i][1]))ans.push_back(i);
    }
    for(int it:ans)cout<<it<<"\n";
    return 0;
}