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;
template <class T>
T smin(T &a, T b) {
  if (a > b) a = b;
  return a;
}
template <class T>
T smax(T &a, T b) {
  if (a < b) a = b;
  return a;
}
const int N = (int)(1e6) + 5;
int p[N];
int q[N];
int main(void) {
  ios_base ::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    int v;
    cin >> v;
    p[v] = i;
    q[i] = v;
  }
  int WAS = 0;
  auto sw = [&](int a, int b) {
    int x = p[a];
    int y = p[b];
    swap(p[a], p[b]);
    swap(q[x], q[y]);
  };
  while (1) {
    ++WAS;
    int ok = 0;
    for (int i = 2; i <= n; ++i)
      for (int j = i; j > 1 && p[j - 1] - p[j] >= k; --j) ok = 1, sw(j, j - 1);
    if (!ok) break;
  }
  assert(WAS == 2);
  for (int i = 1; i <= n; ++i) cout << q[i] << " \n"[i == n];
  return 0;
}
