#include <cstdio>
#include <iostream>
using namespace std;
const int maxn=2e5+50;
int n,k,c;
int suf[maxn<<1];
char S[maxn];
int main()
{
    scanf("%d%d%d",&n,&k,&c);
    scanf("%s",S+1);
    for(int i=n;i>=1;--i)
    {
        if(S[i]=='x') suf[i]=suf[i+1];
        else suf[i]=1+suf[i+c+1];
    }
    int cnt=0;
    for(int i=1,las=-c;i<=n;++i) if(i-las>c)
    {
        if(S[i]=='x') continue;
        if(cnt+suf[i+1]<k) printf("%d\n",i);
        ++cnt;
        las=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 <cstdio>
#include <iostream>
using namespace std;
const int maxn=2e5+50;
int n,k,c;
int suf[maxn<<1];
char S[maxn];
int main()
{
    scanf("%d%d%d",&n,&k,&c);
    scanf("%s",S+1);
    for(int i=n;i>=1;--i)
    {
        if(S[i]=='x') suf[i]=suf[i+1];
        else suf[i]=1+suf[i+c+1];
    }
    int cnt=0;
    for(int i=1,las=-c;i<=n;++i) if(i-las>c)
    {
        if(S[i]=='x') continue;
        if(cnt+suf[i+1]<k) printf("%d\n",i);
        ++cnt;
        las=i;
    }
    return 0;
}
```