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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
const int N = 5e5 + 5;
int a[N], pos[N];
int main() {
  int n = read();
  int k = read();
  for (int i = 1; i <= n; ++i) pos[a[i] = read()] = i;
  if (k == 1) {
    for (int i = 1; i <= n; ++i) printf("%d\n", i);
    return 0;
  }
  if (k == n - 1) {
    if (a[1] == a[n] + 1) swap(a[1], a[n]);
    for (int i = 1; i <= n; ++i) printf("%d\n", a[i]);
    return 0;
  }
  while (1) {
    bool flag = 0;
    for (int i = 1; i <= n; ++i) {
      while (pos[a[i] - 1] - pos[a[i]] >= k) {
        swap(a[pos[a[i]]], a[pos[a[i] - 1]]);
        swap(pos[a[i]], pos[a[i] + 1]);
        flag = 1;
      }
    }
    if (!flag) break;
  }
  for (int i = 1; i <= n; ++i) printf("%d\n", a[i]);
  return 0;
}
