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 FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define whole(x) (x).begin(),(x).end()
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
using ll = long long;
using P = pair<int, int>;
const int mod = 1000000007;

vector<int> to[200005];

int main(){
    int n, k, c;
    cin >> n >> k >> c;
    string s;
    cin >> s;
    vector<bool> work1(n, false);
    vector<bool> work2(n, false);

    int i = 0;
    int rem = k;
    while (i<n && rem>0) {
        if (s[i]=='o') {
            work1[i] = true;
            rem--;
            i += (c+1);
        }
        else {
            i++;
        }
    }
    i = n-1;
    rem = k;
    while (i>=0 && rem>0) {
        if (s[i]=='o') {
            work2[i] = true;
            rem--;
            i -= (c+1);
        }
        else {
            i--;
        }
    }

    
    rep(i, n) {
        if (work1[i] && work2[i]) {
            cout << i+1 << endl;
        }
    }

    return 0;
}