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>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T1, class T2>
ostream& operator<<(ostream& s, pair<T1, T2> P) {
  return s << '<' << P.first << ", " << P.second << '>';
}
template <class T>
ostream& operator<<(ostream& s, vector<T> P) {
  for (int i = 0; i < P.size(); ++i) {
    if (i > 0) {
      s << " ";
    }
    s << P[i];
  }
  return s;
}
int N, K;
vector<int> P, Q;
void mergesort(vector<int>& a, int left, int right) {
  if (right - left <= 1) return;
  int mid = (left + right) / 2;
  mergesort(a, left, mid);
  mergesort(a, mid, right);
  vector<int> pl(right - mid);
  int pointer = mid;
  for (int i = right - 1; i >= mid; --i) {
    int val = a[i];
    while (pointer > left && a[pointer - 1] >= val + K) --pointer;
    pl[i - mid] = pointer;
  }
  vector<int> res;
  int last = left;
  for (int i = 0; i < pl.size(); ++i) {
    for (int j = last; j < pl[i]; ++j) {
      res.push_back(a[j]);
    }
    last = max(last, pl[i]);
    res.push_back(a[i + mid]);
  }
  for (int i = last; i < mid; ++i) res.push_back(a[i]);
  for (int i = 0; i < res.size(); ++i) a[i + left] = res[i];
}
int main() {
  cin >> N >> K;
  P.resize(N);
  Q.resize(N);
  for (int i = 0; i < N; ++i) cin >> P[i], --P[i];
  for (int i = 0; i < N; ++i) Q[P[i]] = i;
  mergesort(Q, 0, N);
  vector<int> res(N);
  for (int i = 0; i < N; ++i) res[Q[i]] = i;
  for (int i = 0; i < N; ++i) cout << res[i] + 1 << endl;
}
