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 = 500005;
int f[N], a[N], l, b[N], tot, ne[N], fi[N], zz[N], in[N], ans[N], n, m;
struct Tree {
  int l, r, num;
} T[N * 4];
void up(int x) {
  if (x == 1) return;
  if (b[f[x]] < b[f[x / 2]]) {
    swap(f[x], f[x / 2]);
    up(x / 2);
  }
}
void down(int x) {
  int i = x;
  if (x * 2 <= l && b[f[x]] > b[f[x * 2]]) i = x * 2;
  if (x * 2 < l && b[f[i]] > b[f[x * 2 + 1]]) i = x * 2 + 1;
  if (i != x) {
    swap(f[x], f[i]);
    down(i);
  }
}
void build(int x, int l, int r) {
  T[x].l = l;
  T[x].r = r;
  T[x].num = n + 1;
  if (T[x].l == T[x].r) return;
  int mid = (l + r) / 2;
  build(x * 2, l, mid);
  build(x * 2 + 1, mid + 1, r);
}
void insert(int x, int y, int z) {
  T[x].num = min(T[x].num, z);
  if (T[x].l == T[x].r) return;
  int mid = (T[x].l + T[x].r) / 2;
  if (y <= mid)
    insert(x * 2, y, z);
  else
    insert(x * 2 + 1, y, z);
}
int find(int x, int l, int r) {
  if (T[x].l > r || l > T[x].r) return n + 1;
  if (T[x].l >= l && T[x].r <= r) return T[x].num;
  return min(find(x * 2, l, r), find(x * 2 + 1, l, r));
}
void jb(int x, int y) {
  ne[++tot] = fi[x];
  fi[x] = tot;
  zz[tot] = y;
  in[y]++;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[a[i]] = i;
  build(1, 1, n);
  for (int i = n; i; i--) {
    int k1 = find(1, b[i] - m + 1, b[i]), k2 = find(1, b[i], b[i] + m - 1);
    if (k1 <= n) jb(i, k1);
    if (k2 <= n) jb(i, k2);
    insert(1, b[i], i);
  }
  for (int i = 1; i <= n; i++)
    if (!in[i]) f[++l] = i, up(l);
  for (int i = 1; i <= n; i++) {
    int k = f[1];
    f[1] = f[l--];
    down(1);
    ans[b[k]] = i;
    for (int j = fi[k]; j; j = ne[j]) {
      in[zz[j]]--;
      if (!in[zz[j]]) f[++l] = zz[j], up(l);
    }
  }
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
