You are given a permutation P_1 ... P_N of the set {1, 2, ..., N}.

You can apply the following operation to this permutation, any number of times (possibly zero):

* Choose two indices i,j (1 ≦ i < j ≦ N), such that j - i ≧ K and |P_i - P_j| = 1. Then, swap the values of P_i and P_j.



Among all permutations that can be obtained by applying this operation to the given permutation, find the lexicographically smallest one.

Constraints

* 2≦N≦500,000
* 1≦K≦N-1
* P is a permutation of the set {1, 2, ..., N}.

Input

The input is given from Standard Input in the following format:


N K
P_1 P_2 ... P_N


Output

Print the lexicographically smallest permutation that can be obtained.

Examples

Input

4 2
4 2 3 1


Output

2
1
4
3


Input

5 1
5 4 3 2 1


Output

1
2
3
4
5


Input

8 3
4 5 7 8 3 1 2 6


Output

1
2
6
7
5
3
4
8
#include<bits/stdc++.h>
#define itr(i, n) for(int i=1;i<=n;++i)
using namespace std;
const int SIZE = 5e5+1;
vector<int> num_arr(SIZE), idx_arr(SIZE);
stack<int> s, t;
int memo[SIZE];
int n, k;
void NIT(){ //Num Index Transform　数とインデックスを入れ替える
    itr(i, n) //両方とも要素を１からｎにする
        idx_arr[num_arr[i]] = i;
}
void INT(){ //Index Num Transform
    itr(i, n)
    {
        num_arr[idx_arr[i]] = i;
    }
    
}
void divide_and_conquer_method(int left, int right){
    if (left >= right) return;
    int median = (left + right) / 2;
    divide_and_conquer_method(left, median);
    divide_and_conquer_method(median+1, right);
    int Min = idx_arr[median];
    s.push(Min);
    t.push(idx_arr[median]);
    for (int i = median-1; i >= left; --i){
        Min = min(Min, idx_arr[i]);
        s.push(Min);
        t.push(idx_arr[i]);
    }
    int j = left;
    for (int i = median+1; i <= right; i++)
    {
        while (!(s.empty()) && s.top() - idx_arr[i] < k)
        {
            memo[j] = t.top();
            ++j;
            s.pop();
            t.pop();
        }
        memo[j] = idx_arr[i];
        j++;
    }
    while (!t.empty())
    {
        memo[j] = t.top();
        ++j;
        t.pop();
        s.pop();
    }
    
    for (int i = left; i <= right; i++){
        idx_arr[i] = memo[i];
    }
}
int main(){
    cin >> n >> k;
    itr(i, n)
        cin >> num_arr[i];
    NIT();
    divide_and_conquer_method(1, n);
    INT();
    itr(i, n){
        cout << num_arr[i] << endl;
    }
}