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 n, nn, k, a[N], b[N], cnt, ans[N];
bool mark[N];
int seg[N << 1];
priority_queue<int> s;
void upd(int p, int val) {
  p += nn, seg[p] = val, p /= 2;
  for (; p > 0; p /= 2) seg[p] = max(seg[p * 2], seg[p * 2 + 1]);
  return;
}
int get(int l, int r) {
  int rtr = -inf;
  for (l += nn, r += nn; l < r; l /= 2, r /= 2) {
    if (l % 2) rtr = max(rtr, seg[l++]);
    if (r % 2) rtr = max(rtr, seg[--r]);
  }
  return rtr;
}
void add(int plc) {
  if (plc >= cnt || plc < 0) return;
  int mx = get(k * plc, k * (plc + 1));
  if (mx < 0) return;
  if (!mark[b[mx]] && get(max(0, b[mx] - k + 1), min(n, b[mx] + k)) == mx)
    s.push(b[mx]), mark[b[mx]] = true;
  return;
}
void makeFind(int plc) {
  add(plc - 1);
  add(plc);
  add(plc + 1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k, nn = n, cnt = (n + k - 1) / k;
  while (__builtin_popcount(nn) != 1) nn++;
  for (int i = 0; i < n; i++) {
    cin >> a[i], a[i]--, b[a[i]] = i;
    upd(i, a[i]);
  }
  for (int i = 0; i < cnt; i++) add(i);
  for (int i = 0; i < n; i++) {
    int ls = s.top();
    s.pop();
    ans[ls] = n - i - 1;
    upd(ls, -inf);
    int plc = ls / k;
    makeFind(plc);
  }
  for (int i = 0; i < n; i++) cout << ans[i] + 1 << "\n";
  return 0;
}
