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 maxn = 500005;
int n, k, x, y, h;
int a[maxn], p[maxn], ans[maxn], r[maxn], firs[maxn];
priority_queue<int, vector<int>, greater<int> > q;
struct edge {
  int too, nex;
} g[maxn];
void top_sort() {
  for (register int i = 1; i <= n; ++i)
    for (register int j = firs[i]; j; j = g[j].nex) r[g[j].too]++;
  for (register int i = 1; i <= n; ++i)
    if (!r[i]) q.push(i);
  int beg, j, tp = n;
  while (q.size()) {
    beg = q.top();
    q.pop();
    ans[tp--] = beg;
    for (register int i = firs[beg]; i; i = g[i].nex) {
      j = g[i].too;
      r[j]--;
      if (!r[j]) q.push(j);
    }
  }
}
void add(int x, int y) {
  g[++h].nex = firs[x];
  firs[x] = h;
  g[h].too = y;
}
int main() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; ++i) scanf("%d", &a[i]), p[a[i]] = i;
  for (register int i = 1; i <= n; ++i) {
    for (register int j = max(1, i - k + 1); j <= i - 1; ++j)
      if (a[i] < a[j]) add(i, j);
    for (register int j = i + 1; j <= min(n, i + k - 1); ++j)
      if (a[i] < a[j]) add(i, j);
  }
  top_sort();
  for (register int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
