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>

using namespace std;

int n, k;
int arr[500100];
vector<int> lis[500100];
map<int,int> mi;

int ans[500100];
int dfs(int here, int s = 1) {
    int sz = 1;
    for (int &there : lis[here]) {
        sz += dfs(there,s+sz-1);
    }
    ans[here]=s+sz-1;
    return sz;
}

int main() {
    int i;

    scanf("%d%d",&n,&k);
    for (i=1;i<=n;i++) scanf("%d",&arr[i]);
    mi[n+1] = 0;
    for (i=1;i<=n;i++) {
        if (i>k) mi.erase(arr[i-k]);
        mi[arr[i]] = i;
        auto it = mi.find(arr[i]);
        it++;
        lis[(*it).second].push_back(i);
    }
    dfs(0);
    for (i=1;i<=n;i++) printf("%d\n",ans[i]);

    return 0;
}
