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>
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
using namespace ::std;
const long long maxn = 5e5 + 500;
const long long inf = 1e9 + 800;
const long long mod = 1e9 + 7;
long long a[maxn];
long long koj[maxn];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    a[i]--;
    koj[a[i]] = i;
  }
  while (1) {
    bool b = 0;
    for (long long i = 0; i + 1 < n; i++) {
      long long v = koj[i];
      long long u = koj[i + 1];
      if (u + k <= v) {
        swap(koj[i], koj[i + 1]);
        swap(a[v], a[u]);
        b = 1;
      }
    }
    if (!b) break;
  }
  for (long long i = 0; i < n; i++) {
    cout << a[i] + 1 << '\n';
    ;
  }
}
