#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<int(n);++i)
#define rep1(i,n) for(int i=1;i<=int(n);++i)

int main(void){
    int n,k,c;
    cin >> n >> k >> c;
    string s;
    cin >> s;
    vector<int> L = {-n}, R = {n*2};
    rep(i,n){
        if (s[i] == 'o' && L.back() + c < i) {
            L.push_back(i);
        }
    }
    rep(ii,n){
        int i = n-1-ii;
        if (s[i] == 'o' && i < R.back() - c) {
            R.push_back(i);
        }
    }
    rep1(i,k){
        if (L[i] == R[k-i+1]) {
            cout << L[i]+1 << 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>
using namespace std;
#define rep(i,n) for(int i=0;i<int(n);++i)
#define rep1(i,n) for(int i=1;i<=int(n);++i)

int main(void){
    int n,k,c;
    cin >> n >> k >> c;
    string s;
    cin >> s;
    vector<int> L = {-n}, R = {n*2};
    rep(i,n){
        if (s[i] == 'o' && L.back() + c < i) {
            L.push_back(i);
        }
    }
    rep(ii,n){
        int i = n-1-ii;
        if (s[i] == 'o' && i < R.back() - c) {
            R.push_back(i);
        }
    }
    rep1(i,k){
        if (L[i] == R[k-i+1]) {
            cout << L[i]+1 << endl;
        }
    }
    return 0;
}

```