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: Accepted)

/*** author: yuji9511 ***/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using lpair = pair<ll, ll>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define rep(i,m,n) for(ll i=(m);i<(n);i++)
#define rrep(i,m,n) for(ll i=(m);i>=(n);i--)
#define printa(x,n) for(ll i=0;i<n;i++){cout<<(x[i])<<" \n"[i==n-1];};
void print() {}
template <class H,class... T>
void print(H&& h, T&&... t){cout<<h<<" \n"[sizeof...(t)==0];print(forward<T>(t)...);}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll N,K,C;
    cin >> N >> K >> C;
    string S;
    cin >> S;
    vector<ll> pos;
    rep(i,0,N){
        if(S[i] == 'o') pos.push_back(i);
    }
    ll cnt1[200010] = {}, cnt2[200010] = {};
    ll cur = pos[0];
    cnt1[pos[0]] = 1;
    ll sz = pos.size();
    rep(i,0,K-1){
        ll val = cur + C + 1;
        ll nxt = lower_bound(pos.begin(), pos.end(), val) - pos.begin();
        cnt1[pos[nxt]] = i+2;
        cur = pos[nxt];
    }
    cur = pos[sz-1];
    cnt2[pos[sz-1]] = K;
    rep(i,0,K-1){
        ll val = cur - C - 1;
        ll nxt = upper_bound(pos.begin(), pos.end(), val) - pos.begin();
        cnt2[pos[nxt-1]] = K-1-i;
        cur = pos[nxt-1];

    }
    // printa(cnt, N);
    rep(i,0,N){
        if(cnt1[i] == cnt2[i] && cnt1[i] != 0){
            print(i+1);
        }
    }

    

}