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 inf = 1e9 + 9;
const int mod = 1e9 + 7;
const long long int MOD = 1e9 + 696969;
const long long INF = 1e18 + 3;
int tab[500100], p[500100], n, k;
const int R = (1 << 19);
pair<int, int> dr[2 * R + 5];
inline void update(int x, int v) {
  x += R - 1;
  dr[x] = make_pair(v, x - R + 1);
  x >>= 1;
  while (x) {
    dr[x] = max(dr[x << 1 | 0], dr[x << 1 | 1]);
    x >>= 1;
  }
}
inline pair<int, int> getMax(int x, int y) {
  x += R - 1;
  y += R - 1;
  pair<int, int> res = make_pair(-1, -1);
  res = max(res, dr[x]);
  if (x == y) return res;
  res = max(res, dr[y]);
  while (x + 1 < y) {
    if (!(x & 1)) res = max(res, dr[x + 1]);
    if (y & 1) res = max(res, dr[y - 1]);
    x >>= 1;
    y >>= 1;
  }
  return res;
}
int DL, odp[500100];
set<int> best;
inline void poprawWiaderko(int bucket) {
  pair<int, int> f = getMax(max(1, bucket * k), min(n, (bucket + 1) * k - 1));
  if (getMax(max(1, f.second - k + 1), min(n, f.second + k - 1)) == f &&
      f.first != 0)
    best.insert(-f.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = (1); i <= (n); ++i) {
    cin >> p[i];
    tab[i] = p[i];
  }
  best.clear();
  for (int i = (1); i <= (n); ++i) update(i, tab[i]);
  for (int i = (0); i <= (n / k); ++i) poprawWiaderko(i);
  int MX = n / k;
  for (int step = (1); step <= (n); ++step) {
    assert(!best.empty());
    int i = -(*best.begin());
    odp[step] = i;
    best.erase(best.begin());
    int bucket = i / k;
    update(i, 0);
    poprawWiaderko(bucket);
    if (bucket != MX) poprawWiaderko(bucket + 1);
    if (bucket != 0) poprawWiaderko(bucket - 1);
  }
  for (int i = n; i > 0; --i) cout << odp[i] << "\n";
}
