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 + 10, INF = 1 << 28;
struct edge {
  int nxt, to;
} e[MAXN << 2];
int head[MAXN], tot;
int n, k, in[MAXN], a[MAXN], pos[MAXN], seg[MAXN << 2];
void add_edge(int x, int y) {
  e[++tot] = (edge){head[x], y};
  head[x] = tot;
  in[y]++;
}
void Update(int val, int k, int l, int r) {
  if (l == r) return;
  seg[k] = min(seg[k], val);
  if (val <= ((l + r) >> 1))
    Update(val, k << 1, l, ((l + r) >> 1));
  else
    Update(val, k << 1 | 1, ((l + r) >> 1) + 1, r);
}
void Query(int &mn, int a, int b, int k, int l, int r) {
  if (a <= l && r <= b) {
    mn = min(mn, seg[k]);
    return;
  }
  if (a <= ((l + r) >> 1)) Query(mn, a, b, k << 1, l, ((l + r) >> 1));
  if (b > ((l + r) >> 1)) Query(mn, a, b, k << 1 | 1, ((l + r) >> 1) + 1, r);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), pos[a[i]] = i;
  fill(seg + 1, seg + (MAXN << 2), INF);
  for (int i = n; i >= 1; i--) {
    int nxt = INF, pre = INF;
    Query(nxt, pos[i], pos[i] + k - 1, 1, 1, n);
    Query(pre, pos[i] - k + 1, pos[i], 1, 1, n);
    if (nxt != INF) add_edge(pos[i], nxt);
    if (pre != INF) add_edge(pos[i], pre);
    Update(pos[i], 1, 1, n);
  }
  priority_queue<int, vector<int>, greater<int> > q;
  for (int i = 1; i <= n; i++)
    if (!in[i]) q.push(i);
  while (!q.empty()) {
    int t = q.top();
    q.pop();
    printf("%d\n", t);
    for (int i = head[t]; i; i = e[i].nxt)
      if (!(--in[e[i].to])) q.push(e[i].to);
  }
  return 0;
}
