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 = 4e5 + 10;
int n, k, a[N], pos[N], deg[N], ans[N];
vector<int> g[N];
int val[N << 2];
void build(int t, int l, int r) {
  val[t] = n + 1;
  if (l == r) return;
  int mid = (l + r) >> 1;
  build((t << 1), l, mid);
  build((t << 1 | 1), mid + 1, r);
}
void insert(int t, int l, int r, int pos, int vl) {
  val[t] = min(val[t], vl);
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (pos <= mid)
    insert((t << 1), l, mid, pos, vl);
  else
    insert((t << 1 | 1), mid + 1, r, pos, vl);
}
int query(int t, int l, int r, int ql, int qr) {
  if (ql > qr) return n + 1;
  if (ql <= l && r <= qr) return val[t];
  int mid = (l + r) >> 1, res = n + 1;
  if (ql <= mid) res = min(res, query((t << 1), l, mid, ql, qr));
  if (qr > mid) res = min(res, query((t << 1 | 1), mid + 1, r, ql, qr));
  return res;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    pos[a[i]] = i;
  }
  build(1, 1, n);
  for (int i = n; i >= 1; i--) {
    int l = query(1, 1, n, max(1, pos[i] - k + 1), pos[i] - 1);
    int r = query(1, 1, n, pos[i] + 1, min(n, pos[i] + k - 1));
    if (r <= n) {
      g[pos[i]].push_back(pos[r]);
      deg[pos[r]]++;
    }
    if (l <= n) {
      g[pos[i]].push_back(pos[l]);
      deg[pos[l]]++;
    }
    insert(1, 1, n, pos[i], i);
  }
  int cur = 0;
  priority_queue<int, vector<int>, greater<int> > q;
  for (int i = 1; i <= n; i++) {
    if (!deg[i]) {
      q.push(i);
    }
  }
  while (q.size()) {
    int u = q.top();
    q.pop();
    ans[u] = ++cur;
    for (auto v : g[u]) {
      if (!(--deg[v])) {
        q.push(v);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}
