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;
int n, k, a[N], pos[N], deg[N];
set<int> S;
vector<int> G[N];
priority_queue<int> q;
void toposort() {
  for (int i = 1; i <= n; i++)
    if (deg[i] == 0) q.push(-i);
  while (!q.empty()) {
    int x = -q.top();
    q.pop();
    printf("%d\n", x);
    for (int i = 0; i < G[x].size(); i++) {
      int y = G[x][i];
      if (--deg[y] == 0) q.push(-y);
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  for (int i = n; i; i--) {
    set<int>::iterator p;
    p = lower_bound(S.begin(), S.end(), pos[i] - k + 1);
    if (p != S.end() && abs(*p - pos[i]) < k)
      G[pos[i]].push_back(*p), deg[*p]++;
    p = lower_bound(S.begin(), S.end(), pos[i]);
    if (p != S.end() && abs(*p - pos[i] < k))
      G[pos[i]].push_back(*p), deg[*p]++;
    S.insert(pos[i]);
  }
  toposort();
  return 0;
}
