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;
template <class T>
void debug(T a, T b) {
  for (; a != b; ++a) cerr << *a << ' ';
  cerr << endl;
}
template <class T>
void chmin(T& a, const T& b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T& a, const T& b) {
  if (a < b) a = b;
}
namespace std {
template <class S, class T>
ostream& operator<<(ostream& out, const pair<S, T>& a) {
  out << '(' << a.first << ',' << a.second << ')';
  return out;
}
}  // namespace std
int n, K;
int ar[500005], br[500005];
int res[500005], cnt, ans[500005];
set<int> alive;
void rec(int v) {
  alive.erase(v);
  auto it = alive.lower_bound(v - K);
  while (it != alive.end() && (*it) < v + K) {
    if (ar[*it] < ar[v]) {
      int v2 = *it;
      rec(v2);
      it = alive.lower_bound(v2);
    } else {
      ++it;
    }
  }
  res[cnt++] = v;
}
int main() {
  cin >> n >> K;
  for (int i = 0; i < (n); ++i) {
    scanf("%d", &ar[i]), --ar[i];
    br[ar[i]] = i;
  }
  for (int i = 0; i < (n); ++i) alive.insert(i);
  for (int i = 0; i < (n); ++i)
    if (alive.count(i)) {
      rec(i);
    }
  for (int i = 0; i < (n); ++i) ans[res[i]] = i;
  for (int i = 0; i < (n); ++i) printf("%d\n", ans[i] + 1);
  return 0;
}
