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 max_N = 5e5 + 21;
int n, K, q[max_N], seg[max_N << 2], pos[max_N], ans[max_N];
bool vis[max_N];
std::priority_queue<int> pq;
inline void update(int x) {
  seg[x] = std::max(seg[((x) << 1)], seg[(((x) << 1) | 1)]);
}
void build(int x, int l, int r) {
  if (l == r) {
    scanf("%d", seg + x), q[seg[x]] = l, pos[l] = x;
    return;
  }
  build(((x) << 1), l, ((l + r) >> 1)),
      build((((x) << 1) | 1), ((l + r) >> 1) + 1, r), update(x);
}
void modify(int x, int v) {
  x = pos[x], seg[x] = v;
  for (x >>= 1; x; x >>= 1) update(x);
}
int ql, qr, qa;
void query(int x, int l, int r) {
  if (ql <= l && r <= qr) {
    qa = std::max(qa, seg[x]);
    return;
  }
  if (ql <= ((l + r) >> 1)) query(((x) << 1), l, ((l + r) >> 1));
  if (qr > ((l + r) >> 1)) query((((x) << 1) | 1), ((l + r) >> 1) + 1, r);
}
void check(int id) {
  if (id < 0 || id > (n - 1) / K) return;
  ql = id * K + 1, qr = std::max((id + 1) * K, n);
  qa = 0, query(1, 1, n);
  if (!qa || vis[qa]) return;
  int tmp = qa;
  ql = std::max(1, q[qa] - K + 1), qr = std::min(n, q[qa] + K - 1);
  qa = 0, query(1, 1, n);
  if (tmp != qa) return;
  vis[qa] = 1, pq.push(q[qa]);
}
int main() {
  scanf("%d%d", &n, &K);
  build(1, 1, n);
  for (int i = 0; i <= (n - 1) / K; ++i) check(i);
  for (int k = n; k; --k) {
    int x = pq.top();
    pq.pop(), ans[x] = k, modify(x, 0);
    int id = (x - 1) / k;
    check(id - 1), check(id), check(id + 1);
  }
  for (int i = 1; i <= n; ++i) printf("%d\n", ans[i]);
  return 0;
}
