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 = 500010;
int gi() {
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') f = f * 10 + ch - 48, ch = getchar();
  return f;
}
int n, k, p[N], a[N];
struct node {
  int mn;
} s[N << 2];
void pushup(int k) { s[k].mn = min(s[k << 1].mn, s[k << 1 | 1].mn); }
void build(int l, int r, int k) {
  if (l == r) {
    s[k].mn = n + 1;
    return;
  }
  int mid = l + r >> 1;
  build(l, mid, k << 1);
  build(mid + 1, r, k << 1 | 1);
  s[k].mn = n + 1;
}
int head[N], cnt, ct;
struct edge {
  int fr, to, next;
  bool operator<(const edge a) const { return to < a.to; }
} e[N << 1];
void adde(int k) {
  e[k].next = head[e[k].fr];
  head[e[k].fr] = k;
}
int query(int L, int R, int l, int r, int k) {
  if (l > r) return n + 1;
  if (L == R) return s[k].mn;
  int mid = L + R >> 1, res = n + 1;
  if (l <= mid) res = min(res, query(L, mid, l, r, k << 1));
  if (r > mid) res = min(res, query(mid + 1, R, l, r, k << 1 | 1));
  return res;
}
void modify(int l, int r, int p, int x, int k) {
  if (l == r) {
    s[k].mn = x;
    return;
  }
  int mid = l + r >> 1;
  if (p <= mid)
    modify(l, mid, p, x, k << 1);
  else
    modify(mid + 1, r, p, x, k << 1 | 1);
  pushup(k);
}
int in[N];
priority_queue<int, vector<int>, greater<int> > q;
int id[N];
void topsort() {
  for (int i = 1; i <= n; i++)
    for (int j = head[i]; j; j = e[j].next) {
      int v = e[j].to;
      in[v]++;
    }
  for (int i = 1; i <= n; i++)
    if (!in[i]) q.push(i);
  while (!q.empty()) {
    int k = q.top();
    q.pop();
    printf("%d\n", k);
    for (int i = head[k]; i; i = e[i].next) {
      int v = e[i].to;
      in[v]--;
      if (!in[v]) q.push(v);
    }
  }
}
int main() {
  n = gi();
  k = gi();
  for (int i = 1; i <= n; i++) p[i] = gi(), a[p[i]] = i;
  build(1, n, 1);
  for (int i = n; i; i--) {
    int x = query(1, n, max(a[i] - k + 1, 1), a[i] - 1, 1);
    if (x <= n) e[++cnt].fr = a[i], e[cnt].to = a[x];
    x = query(1, n, a[i] + 1, min(a[i] + k - 1, n), 1);
    if (x <= n) e[++cnt].fr = a[i], e[cnt].to = a[x];
    modify(1, n, a[i], i, 1);
  }
  sort(e + 1, e + cnt + 1);
  for (int i = 1; i <= cnt; i++) adde(i);
  topsort();
  return 0;
}
