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;
int A[500005];
int pos[500005];
int back[500005];
int n, K;
int main() {
  scanf("%d %d", &n, &K);
  for (int i = 0; i < n; i++) {
    scanf("%d", &A[i]);
    A[i]--;
    pos[A[i]] = i;
  }
  memset(back, -1, sizeof(back));
  set<int> st;
  st.insert(n);
  set<int>::iterator it, its;
  for (int i = n - 1; i >= 0; i--) {
    it = st.lower_bound(pos[i] - K + 1);
    int v = *it;
    back[pos[i]] = back[v];
    back[v] = pos[i];
    while (it != st.end()) {
      int f = *it;
      if (f > pos[i]) break;
      its = it;
      it++;
      st.erase(its);
    }
    st.insert(pos[i]);
  }
  stack<int> ans;
  int now = n;
  while (1) {
    now = back[now];
    if (now == -1) break;
    ans.push(now);
  }
  assert(ans.size() == n);
  while (!ans.empty()) {
    printf("%d\n", ans.top() + 1);
    ans.pop();
  }
  return 0;
}
