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;
set<int> S;
set<int>::iterator it;
int n, k, P[500007], Q[500007];
int head[500007], lt[500007 << 1], nt[500007 << 1], ap = 1, d[500007];
int a[500007], len;
inline int rd() {
  int res = 0;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    ;
  while (c >= '0' && c <= '9') res = res * 10 + c - '0', c = getchar();
  return res;
}
inline void link(int a, int b) {
  lt[++ap] = b, nt[ap] = head[a], head[a] = ap, d[b]++;
}
inline void up(int i) {
  int j = i >> 1, x = a[i];
  while (j && x < a[j]) a[i] = a[j], i = j, j >>= 1;
  a[i] = x;
}
inline void down(int i) {
  int j = i << 1, x = a[i];
  while (j <= len) {
    if (j < len && a[j + 1] < a[j]) j++;
    if (a[j] < x)
      a[i] = a[j], i = j, j <<= 1;
    else
      break;
  }
  a[i] = x;
}
inline void push(int b) {
  a[++len] = b;
  up(len);
}
inline int pop() {
  int b = a[1];
  a[1] = a[len--];
  down(1);
  return b;
}
int main() {
  int i, x;
  n = rd(), k = rd();
  for (i = 1; i <= n; i++) Q[P[i] = rd()] = i;
  for (i = 1; i <= n; i++) {
    if (i > k) S.erase(P[i - k]);
    it = S.insert(P[i]).first;
    if (++it != S.end()) link(i, Q[*it]);
  }
  S.clear();
  for (i = n; i >= 1; i--) {
    if (i + k <= n) S.erase(P[i + k]);
    it = S.insert(P[i]).first;
    if (++it != S.end()) link(i, Q[*it]);
  }
  for (i = 1; i <= n; i++)
    if (!d[i]) push(i);
  while (len) {
    for (i = pop(), x = head[i]; x; x = nt[x])
      if (!--d[lt[x]]) push(lt[x]);
    printf("%d\n", i);
  }
  return 0;
}
