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 p = 1e9 + 7;
struct edge {
  int to, nxxt;
} e[1000002 << 1];
int n, a[1000002], b[1000002], head[1000002], cnt = 1, rd[1000002], s[1000002],
                                              tot, k;
priority_queue<int> q;
inline void ins(int u, int v) {
  e[cnt].to = v;
  e[cnt].nxxt = head[u];
  head[u] = cnt++;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[a[i]] = i;
  for (int i = 1, j = n; i < j; i++, j--) swap(b[i], b[j]);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (abs(b[i] - b[j]) < k) ins(i, j), rd[j]++;
  for (int i = 1; i <= n; i++)
    if (!rd[i]) q.push(i);
  while (!q.empty()) {
    int a1 = q.top();
    q.pop();
    s[++tot] = b[a1];
    for (int i = head[a1]; i; i = e[i].nxxt) {
      int t1 = e[i].to;
      rd[t1]--;
      if (!rd[t1]) q.push(t1);
    }
  }
  for (int i = 1, j = n; i < j; i++, j--) swap(s[i], s[j]);
  for (int i = 1; i <= n; i++) a[s[i]] = i;
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  puts("");
}
