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: Accepted)

#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>
#define pb push_back
#define mpr make_pair
#define pii pair<int, int>
#define pll pair<ll, ll>
#define ll long long
#define ld long double
#define all(arr) arr.begin(), arr.end()
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define fo(i, l, r) for (int i = l; i <= r; i++)
#define INF 1000000001
#define inf1 1000000000000000001
#define mod 1000000007
#define pie 3.14159265358979323846264338327950L
#define N 100005
#define mid(l, r) l + (r - l) / 2
#define vec vector<int>
#define vecl vector<ll>
#define umap unordered_map<ll,ll>
#define yes cout << "Yes" << endl;
#define no cout << "No" << endl;
#define endl "\n"
using namespace std;
 
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};  
int ddx[8]={1,1,0,-1,-1,-1,0,1},ddy[8]={0,1,1,1,0,-1,-1,-1};
ll gcd(ll a,ll b){ if(!a)return b;return gcd(b%a,a);}




 
    
void test_case() {

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

    string s;
    cin>>s;

    vec left(n+1,0), right(n+1, 0);

    int cnt = 0;
    

    for(int i=0; i<n; i++) {
        if(s[i] == 'o') {
            left[cnt] = i;
            cnt++;
            i=i+c;
            if(cnt == k) {
                break;
            }
        }
    }

    for(int i=n-1; i>=0; i--) {
        if(s[i] == 'o') {
            
            cnt--;
            right[cnt] = i;
            i=i-c;
            if(cnt == 0) {
                break;
            }
        }
    }

    for(int i=0; i<k; i++) {
        if(left[i] == right[i]) {
            cout<<left[i]+1<<endl;
        }
    }
    









}
 
 
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cout<<fixed<<setprecision(20);
 
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
 
 
    int t = 1;
    // cin >> t;
    while (t--)
    {
        test_case();
    }
}