#include<bits/stdc++.h>
#define REG register
using namespace std;

int n,k,c;

char d[200005];

int l[200005],r[200005];

inline void Work(){
    scanf("%d %d %d",&n,&k,&c);
    scanf("%s",d+1);
    int lst=0;
    for(REG int i=1;i<=n;++i)
        if(((!lst)||i-lst-1>=c)&&d[i]=='o') l[i]=l[i-1]+1,lst=i;
        else l[i]=l[i-1];
    lst=n+1;
    for(REG int i=n;i;--i)
        if((lst==n+1||lst-i-1>=c)&&d[i]=='o') r[i]=r[i+1]+1,lst=i;
        else r[i]=r[i+1];
    for(REG int i=1;i<=n;++i)
        if(l[i-1]+r[i+1]==k-1) printf("%d ",i);
}

int main(){Work();}
### Prompt

Please formulate 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>
#define REG register
using namespace std;

int n,k,c;

char d[200005];

int l[200005],r[200005];

inline void Work(){
    scanf("%d %d %d",&n,&k,&c);
    scanf("%s",d+1);
    int lst=0;
    for(REG int i=1;i<=n;++i)
        if(((!lst)||i-lst-1>=c)&&d[i]=='o') l[i]=l[i-1]+1,lst=i;
        else l[i]=l[i-1];
    lst=n+1;
    for(REG int i=n;i;--i)
        if((lst==n+1||lst-i-1>=c)&&d[i]=='o') r[i]=r[i+1]+1,lst=i;
        else r[i]=r[i+1];
    for(REG int i=1;i<=n;++i)
        if(l[i-1]+r[i+1]==k-1) printf("%d ",i);
}

int main(){Work();}
```