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 INF = 100000000;
int N, K;
vector<int> p;
void input() {
  cin >> N >> K;
  for (int i = 0; i < N; ++i) {
    int t;
    cin >> t;
    p.push_back(t);
  }
}
void solve() {
  if (K == 1) {
    for (int i = 1; i <= N; ++i) {
      cout << i << endl;
    }
    return;
  }
  priority_queue<int, vector<int>, greater<int> > q;
  vector<pair<int, int> > m;
  int b_end = (N + K - 1) / K;
  for (int b = 0; b < b_end; ++b) {
    int t = INF;
    int i_min = -1;
    bool f = 1;
    {
      int i_begin = b * K;
      int i_end = min(N, i_begin + K);
      for (int i = i_begin; i < i_end; ++i) {
        if (p[i] < t) {
          t = p[i];
          i_min = i;
        }
      }
    }
    {
      int i_begin = max(0, i_min - K + 1);
      int i_end = min(N, i_min + K);
      for (int i = i_begin; i < i_end; ++i) {
        if (p[i] < t) {
          f = 0;
          break;
        }
      }
    }
    m.push_back(pair<int, int>(f, i_min));
    if (f) {
      q.push(i_min);
    }
  }
  vector<int> a;
  a.resize(N);
  vector<bool> u;
  u.resize(N);
  for (int i = 0; i < N; ++i) {
    u[i] = false;
  }
  int l = 1;
  while (!q.empty()) {
    int e = q.top();
    q.pop();
    a[e] = l++;
    u[e] = true;
    int bb = e / K;
    int b_begin = max(0, bb - 1);
    int b_end = min((N + K - 1) / K, bb + 2);
    for (int b = b_begin; b < b_end; ++b) {
      int i_min = -1;
      if (b == bb) {
        int t = INF;
        int i_begin = b * K;
        int i_end = min(N, i_begin + K);
        for (int i = i_begin; i < i_end; ++i) {
          if (!u[i] && p[i] < t) {
            t = p[i];
            i_min = i;
          }
        }
        if (i_min < 0) {
          m[b].first = true;
          m[b].second = i_min;
          continue;
        }
      } else {
        if (m[b].first) {
          continue;
        }
        i_min = m[b].second;
      }
      int f = 1;
      {
        int t = p[i_min];
        int i_begin = max(0, i_min - K + 1);
        int i_end = min(N, i_min + K);
        for (int i = i_begin; i < i_end; ++i) {
          if (!u[i] && p[i] < t) {
            f = 0;
            break;
          }
        }
      }
      m[b] = pair<int, int>(f, i_min);
      if (f) {
        q.push(i_min);
      }
    }
  }
  for (int i = 0; i < N; ++i) {
    cout << a[i] << endl;
  }
}
int main() {
  input();
  solve();
  return 0;
}
