#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int n,c,k,dp1[200005],dp2[200005];
char s[200005];

int main()
{
    scanf("%d%d%d%s",&n,&k,&c,s+1);
    dp1[0]=-0x3f3f3f3f,dp2[k+1]=0x3f3f3f3f;
    for(int i=1,cnt=1;i<=n;i++)
        if(s[i]=='o'&&i>dp1[cnt-1]+c) dp1[cnt++]=i;
    for(int i=n,cnt=k;i>=1;i--)
        if(s[i]=='o'&&i<dp2[cnt+1]-c) dp2[cnt--]=i;
    for(int i=1;i<=k;i++)
        if(dp1[i]==dp2[i]) printf("%d\n",dp1[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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

int n,c,k,dp1[200005],dp2[200005];
char s[200005];

int main()
{
    scanf("%d%d%d%s",&n,&k,&c,s+1);
    dp1[0]=-0x3f3f3f3f,dp2[k+1]=0x3f3f3f3f;
    for(int i=1,cnt=1;i<=n;i++)
        if(s[i]=='o'&&i>dp1[cnt-1]+c) dp1[cnt++]=i;
    for(int i=n,cnt=k;i>=1;i--)
        if(s[i]=='o'&&i<dp2[cnt+1]-c) dp2[cnt--]=i;
    for(int i=1;i<=k;i++)
        if(dp1[i]==dp2[i]) printf("%d\n",dp1[i]);
    return 0;
}
```