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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
vector<int> p;
set<int> t,ans;
int d[200005];
vector<int> v[200005];
int main(){
    int n,k,c,mx=0;
    cin>>n>>k>>c;
    string s;
    cin>>s;s=" "+s;
    for(int i=1;i<s.length();i++){
        if(s[i]=='o'){
            p.push_back(i);
            t.insert(i);
        }
    }
    d[0]=0;
    for(int i=0;i<p.size();i++){
        int x=max(0,p[i]-c-1);
        auto num=t.upper_bound(x);num--;
        d[p[i]]=d[*num]+1;
        mx=max(mx,d[p[i]]);
        v[d[p[i]]].push_back(p[i]);
    }
    if(mx>k||mx<k){
        cout<<endl;
        return 0;
    }
    for(int i=k;i>=2;i--){
        if(v[i].size()==1){
            ans.insert(v[i][0]);
            auto num=upper_bound(v[i-1].begin(),v[i-1].end(),v[i][0]-c-1);
            num--;
            if(num!=v[i-1].begin()) continue;
            else ans.insert(*num);
        }
    }
    for(auto x:ans){
        cout<<x<<endl;
    }
}