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;
int n, k;
int pos[10001];
vector<int> inE[10001];
int deg[10001];
bool v[10001];
int MAIN() {
  memset(deg, 0, sizeof(deg));
  memset(v, 0, sizeof(v));
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int t;
    cin >> t;
    pos[t] = i;
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (pos[i] > pos[j] && pos[i] - pos[j] < k) {
        inE[j].push_back(i);
        deg[i]++;
      }
  vector<int> res;
  for (int i = 1; i <= n; i++) {
    int t = 0, where;
    for (int j = 1; j <= n; j++)
      if (deg[j] == 0 && v[j] == false)
        if (pos[j] > t) {
          t = pos[j];
          where = j;
        }
    res.push_back(t);
    v[where] = true;
    for (int j = 0; j < inE[where].size(); j++) deg[inE[where][j]]--;
  }
  reverse(res.begin(), res.end());
  for (int i = 1; i <= n; i++) pos[res[i - 1]] = i;
  for (int i = 1; i <= n; i++) cout << pos[i] << endl;
  return 0;
}
int main() {
  int start = clock();
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  int ret = MAIN();
  return ret;
}
