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;
int n, m;
struct noshu {
  int out;
  int shu;
  int wei;
  bool operator<(const noshu t) const {
    if (out != t.out)
      return out < t.out;
    else
      return shu < t.shu;
  }
} p[500005];
set<noshu> st1, st2;
set<noshu>::iterator it;
struct nowei {
  int out;
  int shu;
  int wei;
  bool operator<(const nowei t) const {
    if (out != t.out)
      return out < t.out;
    else
      return wei < t.wei;
  }
} tt[500005];
set<nowei> stw;
set<nowei>::iterator itw, itt;
int out[500005];
int head[500005], nxt[500005], cnt, wei[500005];
int ans[500005];
void add(int u, int v) {
  wei[cnt] = v;
  nxt[cnt] = head[u];
  head[u] = cnt++;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].shu);
    p[i].wei = i;
    p[i].out = 0;
  }
  memset(head, -1, sizeof(head));
  cnt = 0;
  st1.clear();
  st2.clear();
  int l, r;
  l = 1;
  r = 1;
  for (int i = 1; i <= n; i++) {
    while (l + m < i) {
      st1.erase(p[l]);
      l++;
    }
    it = st1.lower_bound(p[i]);
    if (it != st1.begin()) {
      it--;
      add(it->wei, i);
      out[i]++;
    }
    st1.insert(p[i]);
    while (r - m < i && r <= n) {
      st2.insert(p[r]);
      r++;
    }
    st2.erase(p[i]);
    it = st2.lower_bound(p[i]);
    if (it != st2.begin()) {
      it--;
      add(it->wei, i);
      out[i]++;
    }
  }
  stw.clear();
  for (int i = 1; i <= n; i++) {
    tt[i].out = out[i];
    tt[i].wei = p[i].wei;
    tt[i].shu = p[i].shu;
    stw.insert(tt[i]);
  }
  while (stw.begin() != stw.end() && stw.begin()->out == 0) {
    itw = stw.begin();
    printf("%d\n", itw->wei);
    for (int i = head[itw->wei]; i != -1; i = nxt[i]) {
      itt = stw.find(tt[wei[i]]);
      nowei nono = *itt;
      nono.out--;
      stw.erase(itt);
      stw.insert(nono);
      tt[wei[i]] = nono;
    }
    stw.erase(itw);
  }
}
