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>
#pragma GCC optimize("O2")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int MAXN = 100010, LOG = 20;
int n, m, k, u, v, x, y, t, a, b, ans;
int P[MAXN];
int deg[MAXN];
vector<int> G[MAXN];
set<pair<int, int>> st;
priority_queue<int, vector<int>, greater<int>> pq;
void add_edge(int u, int v) {
  if (!u || v == n + 1) return;
  G[u].push_back(v);
  deg[v]++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  st.insert({0, 0});
  st.insert({n + 1, n + 1});
  for (int i = 1; i <= n; i++) {
    cin >> P[i];
    if (i > k) st.erase({P[i - k], i - k});
    auto it = st.lower_bound({P[i], 0});
    add_edge(i, it->second);
    it--;
    add_edge(it->second, i);
    st.insert({P[i], i});
  }
  for (int i = 1; i <= n; i++)
    if (!deg[i]) pq.push(i);
  while (pq.size()) {
    int v = pq.top();
    pq.pop();
    cout << v << '\n';
    for (int u : G[v])
      if (!--deg[u]) pq.push(u);
  }
  return 0;
}
