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;
  char ch = getchar();
  while (ch < '0' || '9' < ch) ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return x;
}
const int N = 500009;
int n, k;
int p[N], q[N], deg[N];
set<int> front, back;
vector<int> g[N];
int main() {
  freopen("permutation.in", "r", stdin);
  freopen("permutation.out", "w", stdout);
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) q[p[i] = read()] = i;
  for (int i = 1; i < k; i++) front.insert(p[i]);
  for (int i = 1; i <= n; i++) {
    if (i) front.erase(p[i - 1]);
    if (i + k - 1 <= n) front.insert(p[i + k - 1]);
    back.insert(p[i]);
    if (i - k >= 1) back.erase(p[i - k]);
    set<int>::iterator tmp = front.upper_bound(p[i]);
    if (tmp != front.end()) g[q[(*tmp)]].push_back(i), deg[i]++;
    tmp = back.upper_bound(p[i]);
    if (tmp != back.end()) g[q[(*tmp)]].push_back(i), deg[i]++;
  }
  front.clear();
  for (int i = 1; i <= n; i++)
    if (!deg[i]) front.insert(-i);
  int top = 0;
  while (!front.empty()) {
    int u = -(*front.begin());
    p[++top] = u;
    front.erase(front.begin());
    for (int i = 0, e = g[u].size(); i < e; i++)
      if (--deg[g[u][i]] == 0) front.insert(-g[u][i]);
  }
  for (int i = 1; i <= n; i++) q[p[i]] = n - i + 1;
  for (int i = 1; i <= n; i++) printf("%d\n", q[i]);
  return 0;
}
