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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define QOEINDIEDIS freesteed
#define rep(i,a,n) for (int i=a;i<n;i++)
#define pb push_back
#define fill(x,c) memset(x,c,sizeof(x))
#define fi first
#define se second
#define sz(A) (int)(A.size())
#define all(x) (x).begin(),(x).end()
#define caset int ___T; cin>>___T; for(int cs=1;cs<=___T;cs++)
#define get(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
typedef long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
const ll mod=1e9+7;
const int Max = 1e6+10;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int n,k,c; 
string s;
void solve(){
	cin>>n>>k>>c>>s;
	s = '#' + s;
	VI earliest(k+1,0),latest(k+1,0);
	int i=1,j=1;
	while(i<sz(s)&&j<=k){
		if(s[i]=='o'){
			earliest[j++] = i;
			i+=c+1;
		}else{
			i++;
		}
	}
	i=sz(s)-1,j=k;
	while(i>0&&j>=1){
		if(s[i]=='o'){
			latest[j--] = i;
			i-=c+1;
		}else{
			i--;
		}
	}
//	rep(i,1,k+1)cout<<earliest[i]<<" ";cout<<endl;
//	rep(i,1,k+1)cout<<latest[i]<<" ";cout<<endl;
	rep(i,1,k+1)if(earliest[i]==latest[i])cout<<earliest[i]<<"\n";
}
int main() {
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	//caset{
		solve();
	//}
  	return 0;
}