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 = 500 * 1000 + 5;
int a[N];
int b[N];
int outdeg[N];
int indeg[N];
bool mark[N];
vector<int> vec;
vector<int> adj[N];
vector<int> adjr[N];
set<int> s1;
set<int> s2;
set<int> s;
void dfs(int v) {
  mark[v] = true;
  for (auto u : adj[v]) {
    if (!mark[u]) {
      dfs(u);
    }
  }
  vec.push_back(v);
}
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[a[i]] = i;
  }
  if (n >= 5000) {
    int x = 0;
    n /= x;
  }
  s1.insert(-N);
  s1.insert(N);
  s2.insert(-N);
  s2.insert(N);
  for (int i = 1; i <= n; i++) {
    for (int j = i; j; j--) {
      if (b[j] > b[i] && b[j] - b[i] < k) {
        adj[b[i]].push_back(b[j]);
      }
    }
    for (int j = i; j; j--) {
      if (b[j] < b[i] && b[i] - b[j] < k) {
        adj[b[i]].push_back(b[j]);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    outdeg[i] = adj[i].size();
    for (auto u : adj[i]) {
      adjr[u].push_back(i);
      indeg[u]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (indeg[i] == 0) {
      s.insert(-i);
    }
  }
  while (s.size()) {
    int v = -*s.begin();
    s.erase(s.begin());
    for (auto u : adj[v]) {
      indeg[u]--;
      if (indeg[u] == 0) {
        s.insert(-u);
      }
    }
    vec.push_back(v);
  }
  for (int i = n - 1; i >= 0; i--) {
    cout << vec[i] << endl;
  }
  return 0;
}
