#include <bits/stdc++.h>
#define MAXN 500005
#define ll long long
using namespace std;
int n,m,c,t;
int l[MAXN],r[MAXN];
char s[MAXN];
int main() {
    cin>>n>>m>>c;
    scanf("%s",s+1);
    int now=-c-1,cnt=0;
    for(int i=1;i<=n;i++){
        if(s[i]=='x'||i-now<c+1)continue;
        l[++cnt]=i,now=i;
    }
    now=n+c+1;
    if(cnt>m)return 0;
    t=cnt;
    for(int i=n;i>=1;i--){
        if(s[i]=='x'||now-i<c+1)continue;
        r[cnt--]=i,now=i;
    }
    for(int i=1;i<=t;i++){
        if(l[i]==r[i])cout<<l[i]<<endl;
    }
   return 0;
}
### 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 MAXN 500005
#define ll long long
using namespace std;
int n,m,c,t;
int l[MAXN],r[MAXN];
char s[MAXN];
int main() {
    cin>>n>>m>>c;
    scanf("%s",s+1);
    int now=-c-1,cnt=0;
    for(int i=1;i<=n;i++){
        if(s[i]=='x'||i-now<c+1)continue;
        l[++cnt]=i,now=i;
    }
    now=n+c+1;
    if(cnt>m)return 0;
    t=cnt;
    for(int i=n;i>=1;i--){
        if(s[i]=='x'||now-i<c+1)continue;
        r[cnt--]=i,now=i;
    }
    for(int i=1;i<=t;i++){
        if(l[i]==r[i])cout<<l[i]<<endl;
    }
   return 0;
}
```