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 N = 5e5 + 10, inf = 1e8;
int b[N], a[N], n, k, fen[N], val, seg[N << 2], lazy[N << 2], ans[N];
void add(int p) {
  for (p++; p <= n; p += p & -p) fen[p]++;
}
int getPref(int p) {
  int rtr = 0;
  for (; p > 0; p -= p & -p) rtr += fen[p];
  return rtr;
}
int get(int l, int r) { return getPref(r) - getPref(l); }
void relax(int id) {
  int lc = id * 2, rc = lc + 1;
  seg[lc] += lazy[id], seg[rc] += lazy[id];
  lazy[lc] += lazy[id], lazy[rc] += lazy[id];
  lazy[id] = 0;
}
void addseg(int l, int r, int id = 1, int s = 0, int e = n) {
  if (e <= l || r <= s) return;
  if (l <= s && e <= r) {
    seg[id] += val;
    lazy[id] += val;
    return;
  }
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  relax(id);
  addseg(l, r, lc, s, mid);
  addseg(l, r, rc, mid, e);
  seg[id] = min(seg[lc], seg[rc]);
}
int getseg(int id = 1, int s = 0, int e = n) {
  if (e - s == 1) return s;
  relax(id);
  int mid = (s + e) / 2, lc = id * 2, rc = lc + 1;
  if (seg[rc] == 0) return getseg(rc, mid, e);
  return getseg(lc, s, mid);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> b[i], b[i]--;
  for (int i = 0; i < n; i++) a[b[i]] = i;
  for (int i = n - 1; i >= 0; i--) {
    val = get(max(0, a[i] - k + 1), min(n, a[i] + k));
    addseg(a[i], a[i] + 1);
    add(a[i]);
  }
  for (int i = 0; i < n; i++) {
    int plc = getseg();
    ans[n - i - 1] = plc;
    val = -1;
    addseg(max(0, plc - k + 1), min(n, plc + k));
    val = inf;
    addseg(plc, plc + 1);
  }
  for (int i = 0; i < n; i++) cout << ans[i] + 1 << "\n";
  return 0;
}
