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 = 500010, LOG = 20;
int fenwick[MAXN];
void upd(int i, int x) {
  for (; i < MAXN; i += i & -i) fenwick[i] += x;
}
int get(int i) {
  int ans = 0;
  for (; i; i -= i & -i) ans += fenwick[i];
  return ans;
}
int sch(int x) {
  int i = 0;
  for (int lg = LOG - 1; lg >= 0; --lg) {
    if (fenwick[i + (1 << lg)] < x) i += 1 << lg, x -= fenwick[i];
  }
  return i + 1;
}
int N, K;
int outDeg[MAXN], P[MAXN], I[MAXN];
priority_queue<int> pq;
vector<int> rG[MAXN];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K;
  for (int i = 0; i < N; ++i) {
    cin >> P[i];
    I[P[i]] = i;
    upd(P[i], 1);
    if (i >= K) upd(P[i - K], -1);
    int rnk = get(P[i]);
    if (rnk > 1) {
      int j = I[sch(rnk - 1)];
      rG[i].push_back(j);
      ++outDeg[j];
    }
    if (rnk < min(i + 1, K)) {
      int j = I[sch(rnk + 1)];
      rG[j].push_back(i);
      ++outDeg[i];
    }
  }
  for (int i = 0; i < N; ++i) {
    if (outDeg[i] == 0) {
      pq.emplace(i);
    }
  }
  for (int i = N; i >= 1; --i) {
    int u = pq.top();
    pq.pop();
    P[u] = i;
    for (int v : rG[u]) {
      --outDeg[v];
      if (outDeg[v] == 0) pq.emplace(v);
    }
  }
  for (int i = 0; i < N; ++i) {
    cout << P[i] << '\n';
  }
  return 0;
}
