#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
//#define MOD 998244353
#define MOD 1000000007

int main(){
    int n,k,c;
    scanf("%d %d %d",&n,&k,&c);
    char s[200010];
    scanf("%s",s);
    int a[200010],b[200010],pos=0;
    rep(i,1,n+1)if(s[i-1]=='o')a[pos++]=i,i+=c;
    pos=0;
    for(int i=n;i>0;i--)if(s[i-1]=='o')b[pos++]=i,i-=c;
    rep(i,0,k)if(a[i]==b[k-i-1])printf("%d\n",a[i]);
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

Examples

Input

11 3 2
ooxxxoxxxoo


Output

6


Input

5 2 3
ooxoo


Output

1
5


Input

5 1 0
ooooo


Output




Input

16 4 3
ooxxoxoxxxoxoxxo


Output

11
16

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define INF ((1LL<<62)-(1LL<<31))
//#define MOD 998244353
#define MOD 1000000007

int main(){
    int n,k,c;
    scanf("%d %d %d",&n,&k,&c);
    char s[200010];
    scanf("%s",s);
    int a[200010],b[200010],pos=0;
    rep(i,1,n+1)if(s[i-1]=='o')a[pos++]=i,i+=c;
    pos=0;
    for(int i=n;i>0;i--)if(s[i-1]=='o')b[pos++]=i,i-=c;
    rep(i,0,k)if(a[i]==b[k-i-1])printf("%d\n",a[i]);
}
```