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 = 5e5 + 500;
const int OFF = (1 << 19);
int n, k, P[N], Q[N], T[2 * OFF], sol[N];
inline int miin(int x, int y) {
  if (Q[x] < Q[y]) return x;
  return y;
}
void update(int i, int x) {
  Q[i] = x;
  T[OFF + i] = i;
  for (i = (i + OFF) / 2; i; i /= 2) T[i] = miin(T[2 * i], T[2 * i + 1]);
}
int query(int i, int a, int b, int lo, int hi) {
  if (lo <= a && b <= hi) return T[i];
  if (a > hi || b < lo) return n;
  return miin(query(2 * i, a, (a + b) / 2, lo, hi),
              query(2 * i + 1, (a + b) / 2 + 1, b, lo, hi));
}
inline int lokal(int x) {
  return x == query(1, 0, OFF - 1, max(0, x - k + 1), min(n - 1, x + k - 1));
}
priority_queue<pair<int, int>, vector<pair<int, int> >, less<pair<int, int> > >
    PQ;
inline void buket(int x) {
  if (x < 0 || x * k >= n) return;
  int kan = query(1, 0, OFF - 1, x * k, x * k + k - 1);
  if (Q[kan] == n) return;
  if (lokal(kan)) PQ.push({P[kan], kan});
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", P + i);
    P[i]--;
    update(i, P[i]);
  }
  update(n, n);
  for (int i = 0; i * k < n; i++) buket(i);
  int trn = 0;
  for (; !PQ.empty();) {
    int cur = PQ.top().second;
    PQ.pop();
    if (Q[cur] == n) continue;
    sol[cur] = trn++;
    update(cur, n);
    buket(cur / k);
    buket(cur / k - 1);
    buket(cur / k + 1);
  }
  for (int i = 0; i < n; i++) printf("%d\n", sol[i] + 1);
}
