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)

#include <bits/stdc++.h>
#define rep(i, n) for(int i=0;i<n;i++)
using namespace std;
template<class T>bool chmax(T &a, const T &b) { if (b>a) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
const long long INF = 2LL << 60;
using ll = long long;
using P = pair<ll, ll>;

int main(int argc, char const *argv[]) {
    int n, k, c;
    string s;
    cin >> n >> k >> c >> s;
    vector<int> v(k), r(k);
    int idx=0;
    for(int i=0;i<n&&idx<k;i++){
        if(s[i]=='o'){
            v.at(idx)=i;
            idx++;
            i+=c;
        }
    }
    idx=k-1;
    for(int i=n-1;i>=0&&idx>=0;i--){
        if(s[i]=='o'){
            r.at(idx)=i;
            idx--;
            i-=c;
        }
    }
    rep(i,k){
        if(v.at(i)==r.at(i)){
            cout << v.at(i)+1 << endl;
        }
    }

    return 0;
}