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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
int n, K, a[500000 + 5], b[500000 + 5], head[500000 + 5],
    cnt = 0, in[500000 + 5], c[500000 + 5], d[500000 + 5];
struct edge {
  int to, next;
} e[500000 * 4 + 5];
inline void ins(int f, int t) {
  if (f < t) swap(f, t);
  f = a[f];
  t = a[t];
  ++in[t];
  e[++cnt] = (edge){t, head[f]};
  head[f] = cnt;
}
priority_queue<int> q;
set<int> st;
int main() {
  n = read();
  K = read();
  for (int i = 1; i <= n; ++i) b[a[i] = read()] = i;
  for (int i = 1; i < n + K; ++i) {
    set<int>::iterator it, it2;
    if (i <= n) {
      st.insert(a[i]);
      set<int>::iterator it = st.find(a[i]), it2;
      if (it != st.begin()) ins(a[i], *(--(it2 = it)));
      if ((++(it2 = it)) != st.end()) ins(a[i], *it2);
    }
    if (i >= K) {
      it = st.find(a[i - K + 1]);
      if (it != st.begin()) ins(a[i - K + 1], *(--(it2 = it)));
      if ((++(it2 = it)) != st.end()) ins(a[i - K + 1], *it2);
      st.erase(it);
    }
  }
  for (int i = 1; i <= n; ++i)
    if (!in[i]) q.push(i);
  for (int i = 1; i <= n; ++i) {
    int x = q.top();
    c[n + 1 - i] = x;
    q.pop();
    for (int i = head[x]; i; i = e[i].next)
      if (!--in[e[i].to]) q.push(e[i].to);
  }
  for (int i = 1; i <= n; ++i) d[c[i]] = i;
  for (int i = 1; i <= n; ++i) printf("%d\n", d[i]);
  return 0;
}
