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 MAXN = 5e5 + 5;
int n, K;
int a[MAXN], ans[MAXN];
int p[3] = {12313, 1000000007, 10000007};
int Base[3] = {117, 20020624, 19260717};
struct data {
  int x, y, z;
  data() {}
  data(int _x, int _y, int _z) : x(_x), y(_y), z(_z) {}
  bool operator<(const data &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
};
set<data> vis;
int cal(int id) {
  int res = 0;
  for (int i = 1; i <= n; i++) res = (1ll * res * Base[id] + a[i]) % p[id];
  return (res + p[id]) % p[id];
}
bool check() {
  for (int i = 1; i <= n; i++)
    if (a[i] != ans[i]) return a[i] < ans[i];
  return 0;
}
void print() {
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  exit(0);
}
void dfs() {
  data nh = data(cal(1), cal(2), cal(3));
  if (vis.count(nh)) return;
  vis.insert(nh);
  if (check())
    for (int i = 1; i <= n; i++) ans[i] = a[i];
  for (int i = 1; i <= n; i++)
    for (int j = i + K; j <= n; j++) {
      if (abs(a[i] - a[j]) != 1) continue;
      swap(a[i], a[j]);
      dfs();
      swap(a[i], a[j]);
    }
}
int main() {
  freopen("swap.in", "r", stdin);
  freopen("swap.out", "w", stdout);
  scanf("%d%d", &n, &K);
  if (K == 1) {
    for (int i = 1; i <= n; i++) printf("%d\n", i);
    return 0;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), ans[i] = a[i];
  if (!K) print();
  dfs();
  print();
  fclose(stdin);
  fclose(stdout);
}
