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 = 2100000000;
struct Seg {
  static const int DEPTH = 19;
  static const int N = 1 << DEPTH;
  int val[2 * N];
  Seg() {
    for (int i = 0; i < 2 * N; ++i) val[i] = INF;
  }
  void update(int pos, int v) {
    pos += N;
    val[pos] = v;
    pos >>= 1;
    while (pos) {
      val[pos] = min(val[pos * 2], val[pos * 2 + 1]);
      pos >>= 1;
    }
  }
  int query(int L, int R) {
    L += N, R += N;
    int ret = INF;
    while (L < R) {
      if (L & 1) ret = min(ret, val[L++]);
      if (R & 1) ret = min(ret, val[R--]);
      L >>= 1, R >>= 1;
    }
    return ret;
  }
} seg;
int N, K, p[505050];
bool vis[505050];
int ans[505050], a1;
void decide(int idx) {
  for (;;) {
    int least = seg.query(0, p[idx]);
    if (least >= idx + K) break;
    decide(least);
  }
  ans[idx] = a1++;
  vis[idx] = true;
  seg.update(p[idx], INF);
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 0; i < N; ++i) {
    scanf("%d", p + i);
    --p[i];
    vis[i] = false;
    seg.update(p[i], i);
  }
  a1 = 0;
  for (int i = 0; i < N; ++i)
    if (!vis[i]) decide(i);
  for (int i = 0; i < N; ++i) printf("%d\n", ans[i] + 1);
  return 0;
}
