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 + 5, M = 1 << 19;
int n, m, k, p[N], q[N], i, j, xb, a[N], ans[N], T[M * 2 + 5];
inline void upd(int i) { T[i] = min(T[i << 1], T[i << 1 | 1]); }
inline void mdy(int i) {
  for (T[i += M] = 1 << 25; i >>= 1; upd(i))
    ;
}
inline int ask(int i, int v) {
  for (i += M; !(i & 1) || T[i ^ 1] >= v; i >>= 1)
    ;
  for (i ^= 1; i < M;) i = i << 1 | (T[i << 1 | 1] < v);
  return i - M;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (i = 1; i <= n; ++i) cin >> p[i], q[p[i]] = i;
  for (i = 1; i <= n; ++i) T[M + i] = q[i];
  for (i = M - 1; i; --i) upd(i);
  for (i = 1; i <= n; ++i)
    if (T[M + p[i]] < N) {
      for (j = p[i], xb = 0; a[++xb] = q[j], mdy(j), j = ask(j, q[j] + k);)
        ;
      for (; xb;) ans[a[xb--]] = ++m;
    }
  for (i = 1; i <= n; ++i) cout << ans[i] << '\n';
  return 0;
}
