#include<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MAXN=2e5+10;
int L[MAXN],R[MAXN],n,k,c;
char str[MAXN];
int main(){
    scanf("%d %d %d %s",&n,&k,&c,str+1);
    int cnt=0;
    for(int i=1;i<=n;i++){
        if(str[i]=='o') L[++cnt]=i,i+=c;
        if(cnt==k) break;
    }
    cnt=k+1;
    for(int i=n;i>=1;i--){
        if(str[i]=='o') R[--cnt]=i,i-=c;
        if(cnt==1) break;
    }
    for(int i=1;i<=k;i++){
        if(L[i]==R[i]){
            printf("%d\n",L[i]);
        }
    }
    return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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<iostream>
#include<cstdio>
#include<map>
#include<algorithm>
using namespace std;
typedef long long ll;
const int MAXN=2e5+10;
int L[MAXN],R[MAXN],n,k,c;
char str[MAXN];
int main(){
    scanf("%d %d %d %s",&n,&k,&c,str+1);
    int cnt=0;
    for(int i=1;i<=n;i++){
        if(str[i]=='o') L[++cnt]=i,i+=c;
        if(cnt==k) break;
    }
    cnt=k+1;
    for(int i=n;i>=1;i--){
        if(str[i]=='o') R[--cnt]=i,i-=c;
        if(cnt==1) break;
    }
    for(int i=1;i<=k;i++){
        if(L[i]==R[i]){
            printf("%d\n",L[i]);
        }
    }
    return 0;
}

```