#include<bits/stdc++.h>
using namespace std;
const int N = 2e5+5;
char s[N];
int l[N];
int r[N];
int a[N];
int main()
{
    int n,k,c,cnt = 0;
    scanf("%d %d %d %s",&n,&k,&c,s+1);
    for(int i=1;i<=n;i++)
    {
        if(s[i]=='x')   continue;
        a[++cnt] = i;
        int idx = upper_bound(a+1,a+cnt+1,a[cnt]-c-1)-a-1;
        l[cnt] = l[idx]+1;
    }
    for(int i=cnt;i>=1;i--)
    {
        int idx = lower_bound(a+1,a+cnt+1,a[i]+c+1)-a;
        r[i] = r[idx]+1;
    }
    for(int i=1;i<=cnt;i++)
    {
        if(l[i]+r[i]-1==k&&l[i]!=l[i-1]&&r[i]!=r[i+1])  printf("%d\n",a[i]);
    }
    return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int N = 2e5+5;
char s[N];
int l[N];
int r[N];
int a[N];
int main()
{
    int n,k,c,cnt = 0;
    scanf("%d %d %d %s",&n,&k,&c,s+1);
    for(int i=1;i<=n;i++)
    {
        if(s[i]=='x')   continue;
        a[++cnt] = i;
        int idx = upper_bound(a+1,a+cnt+1,a[cnt]-c-1)-a-1;
        l[cnt] = l[idx]+1;
    }
    for(int i=cnt;i>=1;i--)
    {
        int idx = lower_bound(a+1,a+cnt+1,a[i]+c+1)-a;
        r[i] = r[idx]+1;
    }
    for(int i=1;i<=cnt;i++)
    {
        if(l[i]+r[i]-1==k&&l[i]!=l[i-1]&&r[i]!=r[i+1])  printf("%d\n",a[i]);
    }
    return 0;
}

```