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>
const int N = 4e6 + 10;
int in[N], loc[N], p[N], ans[N], tot, n, k;
std::priority_queue<int, std::vector<int>, std::greater<int> > q;
std::set<int> s;
std::set<int>::iterator it;
int head[N], to[N << 1], Next[N << 1], cnt;
void add(int u, int v) { to[++cnt] = v, Next[cnt] = head[u], head[u] = cnt; }
int abs(int x) { return x > 0 ? x : -x; }
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", p), p[p[0]] = i;
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (abs(p[i] - p[j]) < k) add(p[i], p[j]), ++in[p[j]];
  for (int i = 1; i <= n; i++)
    if (!in[i]) q.push(i);
  while (!q.empty()) {
    int now = q.top();
    ans[now] = ++tot;
    q.pop();
    for (int i = head[now]; i; i = Next[i]) {
      int v = to[i];
      --in[v];
      if (!in[v]) q.push(v);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
