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: Time Limit Exceeded)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include<complex>
#include<unordered_map>
#define deg(x) cout<<#x<<"="<<x<<endl;
#define fd(x) for(int i=0;i<x;i++)
#define fdd(a,x) for(int i=a;i<x;i++)
#define fx(x,n) for(int i=n-1;i>=x;i--)
#define mst(x) memset(x,0,sizeof(x));
#define fi first
#define se second
#define TB ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
#define ok "ok"
#define HH puts("");
#define ls rt<<1
#define rs rt<<1|1
#define pb(a,b) a.push_back(b);
#define FIN  freopen("1.in","r",stdin);
#define FOUT freopen("1.out","w",stdout);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<string,int> pss;
typedef unsigned long long ull;
const ll mod=1e9+7;
const ll inf=1e18;
const long double eps=1e-8;
ll qpow(ll a,ll b,ll m){ ll r=1;a%=m;for(;b;b>>=1){if(b&1)r=r*a%m;a=a*a%m;}return r;}
const double pi=acos(-1);

vector<int> v;

char s[203040];
int n,k,c,len;
bool vis[0];
int f[203040];
vector<int> q;
vector<int> ans;
int main(){
    TB
    cin>>n>>k>>c;
    cin>>s;
    len=strlen(s);
    int i=0,flag=0,flog=0;
    f[0]=(s[0]=='o');
    if(s[0]=='o'){
        flag+=c;
        flog++;
        v.push_back(0);
        q.push_back(0);
    }
    fdd(1,len){
        //f[i]+=f[i-1];
        if(s[i]=='o'){
            //f[i]++;
            //q.push_back(0-i);
            //deg(flag)
            if(i>flag){
                v.push_back(i);
                flag=i+c;
                flog++;
            }
        }
    }
    //sort(q.begin(),q.end());
    //f[len]=f[len-1];
    if(flog>k) return 0;
    flag=len;
    for(int i=len-1;i>=0;i--){
        if(s[i]=='o'){
            if(i<flag){
                vector<int>::iterator it = find(v.begin(), v.end(), i);
                if(it != v.end())
                    ans.push_back(*it);
                flag=i-c;
                flog++;
            }
        }
    }
//    flog=len-1;
//    deg(v[1])deg(flog)
//    deg(v[0])
//    for(int i=v.size()-1;i>=0;i--){
//        if(f[flog]-f[v[i]]==0||flog-v[i]==c+1){
//            ans.push_back(v[i]);
//        }else{
//            int g=0-(flog-c);
//            int j=upper_bound(q.begin(),q.end(),g)-q.begin();
//            flog=-(q[j]);
//        }
//    }
    if(ans.size()){
        for(int i=ans.size()-1;i>=0;i--)
            cout<<ans[i]+1<<'\n';
    }
    return 0;
}