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;
#define int long long
#define pii pair<int,int>
#define fastio() ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define test() int t;cin>>t;while(t--)
#define pb push_back
#define nl cout<<endl
#define MOD 1000000007
#define loop(i,start,end) for(int i=start;i<end;i++)
#define N 100001
#define all(v) v.begin(),v.end()
#define oa(a,n) for(int i=0;i<n;i++)cout<<a[i]<<" ";nl
#define ov(a) for(int i=0;i<a.size();i++)cout<<a[i]+1<<" ";nl
#define vinn(n,v) int n;cin>>n;vector<int>v(n);loop(i,0,n)cin>>v[i];
#define vinm(n,m,v) int n,m;cin>>n>>m;vector<int>v(n);loop(i,0,n)cin>>v[i];
#define debug(x) cerr<<#x<<" : "<<x<<endl;
mt19937 rd(time(0)); // i = rd%n
signed main() 
{

    int n,k,c;
    cin>>n>>k>>c;
    string s;
    cin>>s;
    int ans = 0;
    int cnt = 0;
    vector<int>fr,zr;
    for(int i = 0;i<n;i++)
    {
        if(cnt==k)break;
        if(s[i]=='o')
            fr.pb(i),i+=c,cnt++;
    }
    cnt = 0;
    for(int i = n-1;i>=0;i--)
    {
        if(cnt==k)break;
        if(s[i]=='o')
            zr.pb(i),i-=c,cnt++;
    }
    reverse(all(zr));
    // ov(zr);
    // ov(fr);
    for(int i = 0;i<k;i++)if(zr[i]==fr[i] && zr[i]!=0)cout<<fr[i]+1<<endl;



    return 0;
}