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;
inline int read() {
  int x = 0, ch;
  while (!isdigit(ch = getchar()))
    ;
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x;
}
const int MAXN = 500000;
int a[MAXN + 1];
int d[MAXN + 1];
int to[MAXN + 1][2];
priority_queue<int, vector<int>, greater<int> > q;
set<int> s;
set<int>::iterator it;
int main() {
  int n = read(), k = read();
  for (int i = 1; i <= n; i++) a[read()] = i;
  s.insert(0), s.insert(n + 1);
  for (int i = n; i >= 1; i--) {
    it = s.lower_bound(a[i]);
    int R = *it;
    it--;
    int L = *it;
    if (a[i] - L < k && L) d[L]++, to[a[i]][0] = L;
    if (R - a[i] < k && R <= n) d[R]++, to[a[i]][1] = R;
    s.insert(a[i]);
  }
  int x, u;
  for (int i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  while (!q.empty()) {
    printf("%d\n", x = q.top());
    q.pop();
    u = to[x][0];
    if (u) {
      d[u]--;
      if (!d[u]) q.push(u);
    }
    u = to[x][1];
    if (u) {
      d[u]--;
      if (!d[u]) q.push(u);
    }
  }
  return 0;
}
