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 rep(i,x,y) for(ll i=(x);i<(y);i++)
#define rrep(i,x,y) for(ll i=(ll)(y)-1;i>=(x);i--)
#define all(x) (x).begin(),(x).end()
#ifdef LOCAL
#define dump(x)  cerr << #x << " = " << (x) << endl
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl
#define dassert(...) assert(__VA_ARGS__)
#else
#define dump(x)
#define debug(x)
#define dassert(...)
#endif
//#define int long long

typedef long long ll;
const ll MOD = 1e9 + 7;
const long double EPS = 1e-8;

void solve(long long N, long long K, long long C, std::string S){
    vector<pair<int, int>> ldp(N), rdp(N);
    rep(i,0,N) {
        if (S[i] == 'o') {
            ldp[i] = {i>0 ? ldp[i-1].first + 1 : 1, i};
            rep(j,i+1, min(i+1+C, N)) ldp[j] = ldp[i];
            i += C;
        } else if (i > 0) {
            ldp[i] = ldp[i-1];
        }
    }
    rrep(i,0,N) {
        if (S[i] == 'o') {
            rdp[i] = {i<N-1 ? rdp[i+1].first + 1 : 1, i};
            rep(j,max(i-C,0LL), i) rdp[j] = rdp[i];
            i -= C;
        } else if (i < N-1) {
            rdp[i] = rdp[i+1];
        }
    }
    vector<int> answer;
    rep(i,0,N) {
        if (S[i] == 'x') continue;
        int sum = 0, nextLeft = -1e6, nextRight = 1e6;
        if (i>0) {
            sum += ldp[i-1].first;
            nextLeft = ldp[i-1].second;
        }
        if (i<N-1) {
            sum += rdp[i+1].first;
            nextRight = rdp[i+1].second;
        }
        if (nextRight - nextLeft <= C) sum--;
        if (sum < K) answer.push_back(i+1);
    }
    if (answer.empty()) cout << endl;
    rep(i,0,answer.size()) cout << answer[i] << endl;
}

signed main(){
    // ios_base::sync_with_stdio(false);
    // cin.tie(NULL);
    long long N;
    scanf("%lld",&N);
    long long K;
    scanf("%lld",&K);
    long long C;
    scanf("%lld",&C);
    std::string S;
    std::cin >> S;
    solve(N, K, C, S);
    return 0;
}