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 p[100010], q[100010];
set<int> pre, bac;
vector<int> a[100010];
priority_queue<int, vector<int>, greater<int> > que;
int n, k;
int vis[100010];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i]);
    q[p[i]] = i;
  }
  for (int i = 1; i < k; i++) bac.insert(q[i]);
  set<int>::iterator b, g;
  for (int i = 1; i <= n; i++) {
    b = bac.upper_bound(p[i]);
    g = pre.upper_bound(p[i]);
    if (b != bac.end()) {
      a[i].push_back(q[*b]);
      vis[q[*b]]++;
    }
    if (g != pre.end()) {
      a[i].push_back(q[*g]);
      vis[q[*g]]++;
    }
    pre.insert(p[i]);
    if (i - k + 1 > 0) pre.erase(p[i - k + 1]);
    if (i + k - 1 <= n) bac.insert(p[i + k - 1]);
    bac.erase(p[i]);
  }
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) que.push(i);
  while (!que.empty()) {
    int x = que.top();
    que.pop();
    printf("%d\n", x);
    for (int i = 0; i < a[x].size(); i++) {
      vis[a[x][i]]--;
      if (vis[a[x][i]] == 0) que.push(a[x][i]);
    }
  }
}
