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;
bool chkmin(int &a, int b) { return b < a ? a = b, 1 : 0; }
bool chkmax(int &a, int b) { return b > a ? a = b, 1 : 0; }
inline int read() {
  int x = 0, fh = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fh = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ '0');
  return x * fh;
}
void file() {
  freopen("permutation.in", "r", stdin);
  freopen("permutation.out", "w", stdout);
}
int n, k;
const int N = 501000;
int cnt = 0;
int p[N], ans[N], pos[N];
void input() {
  n = read();
  k = read();
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    p[i] = read();
    pos[p[i]] = i;
  }
  memcpy(ans, p, sizeof(p));
}
inline bool cmp(int *a, int *b) {
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i)
    if (a[i] > b[i])
      return true;
    else if (a[i] < b[i])
      return false;
  return false;
}
void solve() {
  bool flag = true;
  while (flag) {
    flag = false;
    for (int i = (1), _end_ = (int)(n - k); i <= _end_; ++i)
      if (p[i] > 1 && pos[p[i] - 1] - pos[p[i]] >= k) {
        flag = true;
        int v = pos[p[i] - 1];
        swap(pos[p[i]], pos[p[i] - 1]);
        swap(p[i], p[v]);
      }
  }
}
void output() {
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) printf("%d\n", p[i]);
}
int main() {
  for (int i = (1), _end_ = (int)(1); i <= _end_; ++i) {
    input();
    solve();
    output();
  }
  return 0;
}
