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 maxn = 5e5 + 5;
using namespace std;
int q[maxn], p[maxn], k, i, j, n, inf, y, a[maxn], d[maxn];
int l[maxn], r[maxn], tree[maxn << 2], now;
struct ar {
  int x, y;
};
bool operator<(ar a, ar b) { return a.y == b.y ? a.x < b.x : a.y < b.y; }
set<ar> w;
inline int read() {
  int n = 0;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - 48;
  return n;
}
void find(int k, int l, int r, int a, int b) {
  if (now <= tree[k]) return;
  if (l == a && r == b) {
    now = tree[k];
    return;
  }
  int m = (l + r) >> 1;
  if (b <= m)
    find(k << 1, l, m, a, b);
  else if (a > m)
    find(k << 1 | 1, m + 1, r, a, b);
  else
    find(k << 1, l, m, a, m), find(k << 1 | 1, m + 1, r, m + 1, b);
}
void ins(int k, int l, int r, int x) {
  tree[k] = (tree[k] < y ? tree[k] : y);
  if (l == r) return;
  int m = (l + r) >> 1;
  if (x <= m)
    ins(k << 1, l, m, x);
  else
    ins(k << 1 | 1, m + 1, r, x);
}
int main() {
  freopen("permutation.in", "r", stdin);
  freopen("permutation.out", "w", stdout);
  n = read(), k = read();
  for (i = 1; i <= n; i++) p[i] = read(), q[p[i]] = i;
  memset(tree, 127 / 3, sizeof(tree));
  inf = tree[0];
  for (i = n; i >= 1; i--) {
    int x = q[i];
    now = inf, find(1, 1, n, (1 > x - k + 1 ? 1 : x - k + 1), x);
    if (now != inf) l[x] = q[now], d[l[x]]++;
    now = inf, find(1, 1, n, x, (x + k - 1 < n ? x + k - 1 : n));
    if (now != inf) r[x] = q[now], d[r[x]]++;
    y = i, ins(1, 1, n, x);
  }
  for (i = 1; i <= n; i++) w.insert((ar){i, d[i]});
  for (i = 1; i <= n; i++) {
    ar now = *w.begin();
    w.erase(w.begin());
    a[i] = now.x;
    if (i == n) continue;
    if (l[now.x]) {
      int x = l[now.x];
      w.erase(w.find((ar){x, d[x]}));
      d[x]--;
      w.insert((ar){x, d[x]});
    }
    if (r[now.x]) {
      int x = r[now.x];
      w.erase(w.find((ar){x, d[x]}));
      d[x]--;
      w.insert((ar){x, d[x]});
    }
  }
  for (i = 1; i <= n; i++) p[a[i]] = i;
  for (i = 1; i <= n; i++) printf("%d\n", p[i]);
}
