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 N = 5e5 + 7;
int tr[4 * N];
void update(int u, int st, int en, int idx, int x) {
  if (st == en)
    tr[u] = x;
  else {
    int mid = (st + en) / 2;
    if (idx <= mid)
      update(2 * u, st, mid, idx, x);
    else
      update(2 * u + 1, mid + 1, en, idx, x);
    tr[u] = max(tr[2 * u], tr[2 * u + 1]);
  }
}
int query(int u, int st, int en, int l, int r) {
  if (r < st || en < l)
    return 0;
  else if (l <= st && en <= r)
    return tr[u];
  else {
    int mid = (st + en) / 2;
    return max(query(2 * u, st, mid, l, r),
               query(2 * u + 1, mid + 1, en, l, r));
  }
}
int a[N], b[N], out[N];
vector<int> in[N];
void add(int u, int v) {
  out[u]++;
  in[v].push_back(u);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    int l1 = query(1, 1, n, b[i] - k + 1, b[i]);
    int l2 = query(1, 1, n, b[i], b[i] + k - 1);
    if (l1) add(b[l1], b[i]);
    if (l2) add(b[l2], b[i]);
    update(1, 1, n, b[i], i);
  }
  vector<int> ans;
  priority_queue<int> pq;
  for (int i = 1; i <= n; i++)
    if (out[i] == 0) pq.push(i);
  while (pq.size()) {
    int u = pq.top();
    pq.pop();
    ans.push_back(u);
    for (int v : in[u]) {
      out[v]--;
      if (out[v] == 0) pq.push(v);
    }
  }
  reverse(ans.begin(), ans.end());
  for (int x : ans) cout << x << "\n";
}
