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 read() {
  int x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f ? x : -x;
}
const int maxn = 5e5 + 10, maxm = maxn << 1;
int e, n, K;
int to[maxm], nxt[maxm], head[maxn], d[maxn];
int Nxt[maxn], pre[maxn], p[maxn], q[maxn], Ans[maxn];
void add(int x, int y) {
  to[++e] = y;
  nxt[e] = head[x];
  head[x] = e;
  ++d[y];
}
void topsort() {
  priority_queue<int, vector<int>, greater<int> > Q;
  for (int i = 1; i <= n; ++i)
    if (!d[i]) Q.push(i);
  int tot = 0;
  while (!Q.empty()) {
    int u = Q.top();
    Q.pop();
    Ans[u] = ++tot;
    for (int i = head[u]; i; i = nxt[i]) {
      int v = to[i];
      --d[v];
      if (!d[v]) Q.push(v);
    }
  }
  for (int i = 1; i <= tot; ++i) printf("%d\n", Ans[i]);
  cout << endl;
}
int main() {
  n = read(), K = read();
  for (int i = 1; i <= n; ++i) p[i] = read(), q[p[i]] = i;
  memset(Nxt, 0x3f, sizeof(Nxt));
  set<int> S;
  for (int i = 1; i <= n; ++i) {
    if (i >= K + 1) S.erase(S.lower_bound(p[i - K]));
    set<int>::iterator it = S.lower_bound(p[i]);
    if (it != S.begin()) add(q[*(--it)], i), ++it;
    if (it != S.end()) add(i, q[*it]);
    S.insert(p[i]);
  }
  S.clear();
  for (int i = n; i >= 1; --i) {
    if (n - i >= K) S.erase(S.lower_bound(p[i + K]));
    set<int>::iterator it = S.lower_bound(p[i]);
    if (it != S.begin()) add(q[*(--it)], i), ++it;
    if (it != S.end()) add(i, q[*it]);
    S.insert(p[i]);
  }
  for (int i = 1; i <= n; ++i) {
  }
  topsort();
  return 0;
}
