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 long double Pi = 3.14159265359;
const long long MOD = 1000 * 1000 * 1000 + 7;
const long long N = 5e5 + 10;
const long long INF = 1e18;
const long long LOG = 20;
long long n, a[N], koj[N], k;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    koj[a[i]] = i;
  }
  long long f = 1;
  long long last = 0;
  long long fir = n - 1;
  while (f) {
    f = 0;
    long long last1 = 0;
    for (int i = min(fir, n - 1); i >= 1; i--) {
      if (koj[i + 1] < koj[i] && koj[i] - koj[i + 1] >= k) {
        a[koj[i + 1]] = i;
        a[koj[i]] = i + 1;
        swap(koj[i], koj[i + 1]);
        last1 = i;
        f = 1;
        fir = max(fir, (long long)i);
      } else {
        if (last > i) i = 0;
      }
    }
    fir++;
    last = last1;
  }
  for (int i = 1; i <= n; i++) cout << a[i] << '\n';
  return 0;
}
