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: Runtime Error)

#include <vector>
#include <map>
#include <set>
#include <sstream>
#include <iostream>
#include <limits>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<pair<int, int>> vii;
typedef stringstream ss;
const int mii = numeric_limits<int>::min();
const int mai = numeric_limits<int>::max();
const ll mil = numeric_limits<long long>::min();
const ll mal = numeric_limits<long long>::max();

#define REP(i, a, b) for(int i=a; i<=b; ++i)
#define RREP(i, a, b) for(int i=a; i>=b; --i)
#define PB(x) push_back(x)

int main(int argc, char **argv) {

    int n, k, c;
    string s;
    cin >> n >> k >> c;
    cin >> s;

    vi l(k);
    vi r(k);

    int cc = 0;

    int kc = 0;
    REP(i, 0, n-1 && kc < k) {
        if(cc > 0) {
            cc--;
            continue;
        }

        if(s[i] == 'o') {
            l[kc] = i;
            ++kc;
            cc = c;
        }

    }

    kc = k-1;
    cc=0;
    RREP(i, n-1, 0 && kc < k) {
        if(cc > 0) {
            cc--;
            continue;
        }

        if(s[i] == 'o') {
            r[kc] = i;
            kc--;
            cc = c;
        }
    }

    REP(i, 0, k-1) {
        if(l[i] == r[i])
            cout << l[i]+1 << endl;
    }
}