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 head[500050], to[500050 << 1], nxt[500050 << 1], cnt, n, a[500050],
    p[500050], du[500050], K, Q[500050], ans[500050];
set<int> S, T;
priority_queue<int> q;
inline void add(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt;
  du[v]++;
}
int main() {
  scanf("%d%d", &n, &K);
  int i, x, lq = 0;
  for (i = 1; i <= n; i++) scanf("%d", &a[i]), p[a[i]] = i;
  for (i = 1; i <= K; i++) S.insert(a[i]);
  for (i = 1; i <= n; i++) {
    set<int>::iterator it;
    it = S.upper_bound(a[i]);
    if (it != S.end()) add(p[*it], i);
    it = T.upper_bound(a[i]);
    if (it != T.end()) add(p[*it], i);
    T.insert(a[i]);
    if (i - K + 1 >= 1) T.erase(a[i - K + 1]);
    if (i + K <= n) S.insert(a[i + K]);
    S.erase(a[i]);
  }
  for (i = 1; i <= n; i++)
    if (!du[i]) q.push(i);
  while (!q.empty()) {
    x = q.top();
    q.pop();
    Q[++lq] = x;
    for (i = head[x]; i; i = nxt[i]) {
      if ((--du[to[i]]) == 0) q.push(to[i]);
    }
  }
  for (i = n; i; i--) printf("%d\n", Q[i]);
}
