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 + 5;
set<int> s;
priority_queue<int, vector<int>, greater<int> > st;
int n, k;
int A[maxn], ans[maxn], ind[maxn];
int tot;
int h[maxn], last[maxn << 1], to[maxn << 1];
void add(int x, int y) {
  to[++tot] = y;
  last[tot] = h[x];
  h[x] = tot;
  ++ind[y];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    int a = 0;
    scanf("%d", &a);
    A[a] = i;
  }
  for (int i = n; i >= 1; --i) {
    set<int>::iterator a = s.lower_bound(A[i] - k + 1);
    set<int>::iterator b = s.lower_bound(A[i] + k - 1);
    if (b != s.begin() && b != s.end()) --b;
    if (a != s.end() && *a <= A[i] + k - 1) {
      add(A[i], *a);
    }
    if (b != s.begin() && *b >= A[i] - k + 1) {
      add(A[i], *b);
    }
    s.insert(A[i]);
  }
  for (int i = 1; i <= n; ++i)
    if (!ind[i]) st.push(i);
  int cnt = 0;
  while (st.size()) {
    int u = st.top();
    st.pop();
    ans[u] = ++cnt;
    for (int i = h[u]; i; i = last[i]) {
      int v = to[i];
      if (!--ind[v]) st.push(v);
    }
  }
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
