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;
template<typename T1,typename T2> bool chmin(T1 &a,T2 b){if(a<=b)return 0; a=b; return 1;}
template<typename T1,typename T2> bool chmax(T1 &a,T2 b){if(a>=b)return 0; a=b; return 1;}




signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(20);

    int n,k,c;
    cin>>n>>k>>c;
    string s;
    cin>>s;
    if(c==0){
        int cnt = 0;
        for(int i=0;i<n;i++){
            if(s[i]='o')cnt++;
        }
        if(cnt == k){
            for(int i=0;i<n;i++){
                if(s[i]=='o'){
                    cout << i +1 << endl;
                }

            }
        }
        return 0;
    }
    vector<int> ans;
    int a[n+1]={};
    int r = 0;
    for(int i=0;i<n;i++){
        if(i) chmax(a[i],a[i-1]);
        if(s[i] == 'o' && i >= r){
            chmax(a[min(n,i+c+1)],a[i]+1);
            r = i+c+1;
        }
    }
    //for(int i=0;i<=n;i++) cout << a[i];
    //cout << endl;
    chmax(a[n],a[n-1]);
    if(a[n] == k){
        int r = n-1;
        for(int p=k-1;p>=0;p--){
            int cnt = 0;
            int v = -1;
            for(int i=r;i>=0;i--){
                if(a[i] != p){
                    r = i;
                    break;
                }
                if(s[i] == 'o'){
                    cnt++;
                    v = i+1;
                }
            }
            //cerr << p << " " << cnt << endl;
            if(cnt == 1){
                ans.push_back(v);
                r = min(r,v-c-2);
            }
        }
    }
    if(ans.size())sort(ans.begin(),ans.end());
    for(auto i:ans){
        cout << i << "\n";
    }

}