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 <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <queue>
#include <string>
#include <map>
#include <set>
#include <stack>
#include <tuple>
#include <deque>
#include <numeric>
#include <bitset>
#include <iomanip>
#include <cassert>
#include <chrono>
#include <random>
#include <limits>
#include <iterator>
#include <functional>
#include <sstream>
#include <complex>
using namespace std;

typedef long long ll;
typedef uint64_t ull;
typedef pair<int, int> P;
constexpr double EPS = 1e-9;
constexpr int INF = 1001001001;
constexpr int mod = 1000000007;
// constexpr int mod = 998244353;

template<class T>
inline bool chmax(T& x, T y){
    if(x < y){
        x = y;
        return true;
    }
    return false;
}
template<class T>
inline bool chmin(T& x, T y){
    if(x > y){
        x = y;
        return true;
    }
    return false;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, k, c;
    string s;
    cin >> n >> k >> c >> s;
    vector<int> L(n, 0), R(n, k + 1), pos;
    for(int i = 0; i < n; ++i){
        if(s[i] == 'o') pos.emplace_back(i);
    }
    L[pos[0]] = 1, R[pos.back()] = k;
    int cur = 0;
    for(int i = 0; i < n; ++i){
        if(s[i] == 'x') continue;
        if(L[i] > cur){
            cur = L[i];
            auto it = upper_bound(pos.begin(), pos.end(), i + c);
            if(it != pos.end()){
                L[*it] = min(k, cur + 1);
            }
        }
        else{
            L[i] = cur;
        }
    }
    cur = k + 1;
    for(int i = n - 1; i >= 0; --i){
        if(s[i] == 'x') continue;
        if(R[i] < cur){
            cur = R[i];
            auto it = lower_bound(pos.begin(), pos.end(), i - c);
            if(it != pos.begin()){
                --it;
                R[*it] = max(1, cur - 1);
            }
        }
        else{
            R[i] = cur;
        }
    }
    vector<int> cntL(k + 1, 0), cntR(k + 1, 0);
    for(int i = 0; i < n; ++i){
        if(s[i] == 'x') continue;
        ++cntL[L[i]];
        ++cntR[R[i]];
    }
    set<int> candidates;
    for(int i = 1; i <= k; ++i){
        if(cntL[i] == 1 || cntR[i] == 1)    candidates.insert(i);
    }
    for(int i = 0; i < n; ++i){
        if(s[i] == 'x') continue;
        if(L[i] != R[i])    continue;
        if(candidates.find(L[i]) != candidates.end()){
            cout << i + 1 << '\n';
        }
    }
}