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 MOD = 1e9 + 7;
const int SIZE = 1e6 + 10;
int a[SIZE], p[SIZE], d[SIZE];
set<int> D, U, H;
vector<int> e[SIZE], an;
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  for (int i = (1); i < (N + 1); ++i) {
    scanf("%d", &(a[i]));
    p[a[i]] = i;
  }
  int rr = 1;
  U.insert(N + 1);
  U.insert(0);
  D.insert(N + 1);
  D.insert(0);
  for (int i = (1); i < (N + 1); ++i) {
    for (; rr < min(N + 1, i + K); rr++) U.insert(p[i]);
    U.erase(p[i]);
    if (i - K > 0) D.erase(p[i - K]);
    set<int>::iterator it = U.lower_bound(p[i]);
    if (*it != N + 1) d[i]++, e[a[*it]].push_back(i);
    it--;
    if (*it != 0) d[a[*it]]++, e[i].push_back(a[*it]);
    it = D.lower_bound(p[i]);
    if (*it != N + 1) d[i]++, e[a[*it]].push_back(i);
    it--;
    if (*it != 0) d[a[*it]]++, e[i].push_back(a[*it]);
    D.insert(p[i]);
  }
  for (int i = (1); i < (N + 1); ++i)
    if (d[i] == 0) H.insert(i);
  for (int kk = 0; kk < (N); ++kk) {
    int x = *H.rbegin();
    H.erase(x);
    an.push_back(x);
    for (int i = 0; i < (((int)(e[x]).size())); ++i) {
      d[e[x][i]]--;
      if (!d[e[x][i]]) H.insert(e[x][i]);
    }
  }
  for (int i = N - 1; i >= 0; i--) printf("%d%c", an[i], i ? ' ' : '\n');
  return 0;
}
