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 MAXN = 5 * 100 * 1000 + 17;
int n, k, a[MAXN], p[MAXN];
set<pair<int, int> > s;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i], p[a[i]] = i;
  for (int i = 1; i < n; i++) s.insert({p[i] - p[i + 1], i});
  int cnt = 0;
  while ((s.rbegin()->first) >= k) {
    pair<int, int> t = *s.rbegin();
    int i = t.second;
    s.erase({p[i] - p[i + 1], i});
    s.erase({p[i - 1] - p[i], i - 1});
    s.erase({p[i + 1] - p[i + 2], i + 1});
    swap(p[i], p[i + 1]);
    s.insert({p[i] - p[i + 1], i});
    if (i != 1) s.insert({p[i - 1] - p[i], i - 1});
    if (i != 4) s.insert({p[i + 1] - p[i + 2], i + 1});
    cnt++;
  }
  cerr << "# " << cnt << endl;
  for (int i = 1; i <= n; i++) a[p[i]] = i;
  for (int i = 1; i <= n; i++) cout << a[i] << '\n';
  return 0;
}
