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 = 5e5 + 20;
int d[maxn], a[maxn], n, mx[maxn * 4];
vector<int> adj[maxn];
void update(int p, int val, int s = 0, int e = n, int v = 1) {
  if (e - s < 2) {
    mx[v] = val;
    return;
  }
  int m = (s + e) / 2;
  if (p < m)
    update(p, val, s, m, 2 * v);
  else
    update(p, val, m, e, 2 * v + 1);
  mx[v] = max(mx[2 * v], mx[2 * v + 1]);
}
int get(int l, int r, int s = 0, int e = n, int v = 1) {
  if (l <= s && e <= r) return mx[v];
  if (r <= s || e <= l) return -1;
  int m = (s + e) / 2;
  return max(get(l, r, s, m, 2 * v), get(l, r, m, e, 2 * v + 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  memset(mx, -1, sizeof mx);
  int k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    x--;
    a[x] = i;
  }
  for (int i = 0; i < n; i++) {
    int v = get(a[i] - k + 1, a[i]);
    if (v >= 0) adj[a[v]].push_back(a[i]), d[a[i]]++;
    v = get(a[i], a[i] + k);
    if (v >= 0) adj[a[v]].push_back(a[i]), d[a[i]]++;
    update(a[i], i);
  }
  set<int> st;
  for (int i = 0; i < n; i++)
    if (!d[i]) st.insert(i);
  for (int _ = 0; _ < n; _++) {
    assert(!st.empty());
    int v = *st.begin();
    st.erase(v);
    for (auto u : adj[v]) {
      d[u]--;
      if (!d[u]) st.insert(u);
    }
    cout << v + 1 << endl;
  }
}
