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 long long N = 1e5 + 30, Mod = 1e9 + 7;
const long long SQ = 330;
long long n, k;
long long d[N];
long long a[N], q[N];
set<pair<long long, long long> > s;
vector<long long> G[N];
void Add(long long v, long long u) {
  if (!~v || u > n) return;
  G[v].push_back(u);
  d[u]++;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> k;
  s.insert({-1, -1});
  s.insert({n + n, n + n});
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (i > k) s.erase({a[i - k], i - k});
    auto it = *s.lower_bound({a[i], -1});
    long long x = it.second;
    Add(i, x);
    auto itt = s.lower_bound({a[i], -1});
    itt--;
    x = itt->second;
    Add(x, i);
    s.insert({a[i], i});
  }
  set<long long> Q;
  for (int i = 1; i <= n; i++)
    if (!d[i]) Q.insert(i);
  for (int i = 1; i <= n; i++) {
    long long x = *Q.begin();
    Q.erase(x);
    q[x] = i;
    for (auto u : G[x]) {
      d[u]--;
      if (!d[u]) Q.insert(u);
    }
  }
  for (int i = 1; i <= n; i++) cout << q[i] << '\n';
  return (0);
}
