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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;

#define REP(i, n, m) for(ll i=n; i<(ll)m; ++i)
#define IREP(i, n, m) for(ll i=n-1; i>=m; --i)
#define rep(i, n) REP(i, 0, n)
#define irep(i, n) IREP(i, n, 0)
#define all(v) v.begin(), v.end()
#define vprint(v) for(auto e:v){cout<<e<<" ";};cout<<endl;
#define vvprint(vv) for(auto v:vv)vprint(v);

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

    ll N, K, C;
    cin >> N >> K >> C;
    string S;
    cin >> S;

    vll l(N, 0), r(N, 0), nl(N, 0), nr(N, 0);
    ll count = 0, last = 0;
    rep(i, N){
        l[i] = count;
        nl[i] = last;
        if(S[i]=='x') continue;
        l[i] = ++count;
        nl[i] = last = i;
        rep(j, C){
            ++i;
            if(i>=N) break;
            l[i] = count;
            nl[i] = last;
        }
    }
    count = 0;
    last = N-1;
    irep(i, N){
        r[i] = count;
        nr[i] = last;
        if(S[i]=='x') continue;
        r[i] = ++count;
        nr[i] = last = i;
        rep(j, C){
            --i;
            if(i<0) break;
            r[i] = count;
            nr[i] = last;
        }
    }

    rep(i, N){
        if(S[i]=='x') continue;
        if(i==0){
            ll t = i+1<N? r[i+1] : 0;
            if(t<K) cout << i+1 << endl;
        }
        else if(i==N-1){
            ll t = i-1>=0? l[i-1] : 0;
            if(t<K) cout << i+1 << endl;
        }
        else{
            ll n1 = nl[i-1];
            ll n2 = nr[i+1];
            ll t = l[n1] + r[n2];
            if(t<K) cout << i+1 << endl;
            else if(t==K && n2-n1<=C) cout << i+1 << endl;
        }
    }
}