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 + 7;
int n, k;
int p[N], q[N];
int seg[N * 15];
bool mrk[N];
int cpk;
set<int> s;
void upd(int p, int val) {
  p += (1 << 19);
  seg[p] = val;
  p /= 2;
  while (p) {
    seg[p] = max(seg[p * 2], seg[p * 2 + 1]);
    p /= 2;
  }
}
int get(int l, int r) {
  l += (1 << 19);
  r += (1 << 19);
  int res = max(seg[l], seg[r]);
  while (l / 2 != r / 2) {
    if (l % 2 == 0) res = max(res, seg[l + 1]);
    if (r % 2 == 1) res = max(res, seg[r - 1]);
    l /= 2;
    r /= 2;
  }
  return res;
}
void get(int pk) {
  int l = pk * k;
  int r = min(n, pk * k + k - 1);
  int mx = get(l, r);
  if (mx < 0 || mrk[q[mx]]) return;
  if (get(q[mx] - k + 1, q[mx] + k - 1) == mx) {
    s.insert(q[mx]);
    mrk[q[mx]] = 1;
  }
}
int res[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  cpk = (n + k - 1) / k;
  for (int i = 0; i < n; i++) {
    cin >> p[i];
    p[i]--;
    q[p[i]] = i;
    upd(i, p[i]);
  }
  for (int i = 0; i < cpk; i++) get(i);
  for (int i = 1; i <= n; i++) {
    int p = *s.rbegin();
    s.erase(p);
    res[p + 1] = n - i + 1;
    upd(p, -1);
    for (int i = (p / k) - 1; i <= (p / k) + 1; i++)
      if (0 <= i && i <= cpk) get(i);
  }
  for (int i = 1; i <= n; i++) cout << res[i] << '\n';
  return 0;
}
