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;

const int N = 5e5 + 5;

int n, k, top;
int p[N], q[N];

map <int, int> M;
list <int> L[N];

int main() {
    cin >> n >> k;
    M[0] = 0;
    for(int i = 1; i <= n; i++) scanf("%d", &p[i]), q[p[i]] = i;
    for(int i = 1; i < k; i++) M[p[i]] = i;
    for(int i = 1; i <= n; i++) L[i].push_back(i);
    for(int i = 1; i <= n; i++) {
        if(i + k - 1 <= n) M[p[i + k - 1]] = i + k - 1;
        auto it = -- M.find(p[i]);
        L[it -> second].splice(L[it -> second].end(), L[i]);
        M.erase(p[i]);
    }
    for(auto x : L[0]) p[++top] = x;
    for(int i = 1; i <= n; i++) q[p[i]] = i;
    for(int i = 1; i <= n; i++) printf("%d\n", q[i]);
    return 0;
}