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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
int const N = 5e5 + 20;
int n, k, a[N], res[N];
set<int> st;
pii seg[N << 2];
void modify(int p, int val, int v = 1, int s = 0, int e = n) {
  if (s + 1 == e) {
    seg[v] = pii(val, s);
    return;
  }
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  if (p < mid)
    modify(p, val, lc, s, mid);
  else
    modify(p, val, rc, mid, e);
  seg[v] = max(seg[lc], seg[rc]);
}
pii get(int l, int r, int v = 1, int s = 0, int e = n) {
  if (s >= r || e <= l) return pii(-1, 0);
  if (s >= l && e <= r) return seg[v];
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  return max(get(l, r, lc, s, mid), get(l, r, rc, mid, e));
}
inline void rel(int b) {
  int xl = b * k, xr = min(n, xl + k);
  pii p = get(xl, xr);
  int index = p.second;
  if (a[index] && p == get(max(0, index - k + 1), min(n, index + k)))
    st.insert(index);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) modify(i, a[i]);
  for (int i = 0; i < n; i += k) rel(i / k);
  for (int i = 0; i < n; i++) {
    int index = *st.rbegin();
    res[index] = n - i;
    st.erase(index);
    a[index] = 0;
    modify(index, a[index]);
    rel(index / k);
    if (index - k >= 0) rel(index / k - 1);
    if (index + k < n) rel(index / k + 1);
  }
  for (int i = 0; i < n; i++) cout << res[i] << '\n';
}
