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;
int n, k;
int val[510000];
int sit[510000], minn[510000];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline void swap(int a, int b) {
  int tmp = val[a];
  val[a] = val[b];
  val[b] = tmp;
  tmp = sit[val[a]];
  sit[val[a]] = sit[val[b]];
  sit[val[b]] = tmp;
}
inline void check() {
  for (register int i = 1; i <= n; i++) {
    if (minn[i] < val[i])
      return;
    else if (minn[i] > val[i]) {
      memcpy(minn, val, sizeof(val));
      return;
    }
  }
}
void work1() {
  for (register int i = 1; i <= 10000000; i++) {
    int a = rand() % n + 1, b = i % 2;
    if ((b == 0 && a > 1) || (a == n)) {
      if (abs(sit[a] - sit[a - 1]) >= k) {
        swap(sit[a], sit[a - 1]);
        check();
      }
    } else if ((b == 1 && a < n) || (a == 1)) {
      if (abs(sit[a] - sit[a + 1]) >= k) {
        swap(sit[a], sit[a + 1]);
        check();
      }
    }
  }
  for (register int i = 1; i <= n; i++) printf("%d\n", minn[i]);
}
void work2() {
  for (register int i = 1; i <= n; i++) printf("%d\n", i);
}
void work3() {
  if (val[1] = val[n] + 1) swap(1, n);
  for (register int i = 1; i <= n; i++) printf("%d\n", val[i]);
}
int main() {
  memset(minn, 999999, sizeof(minn));
  cin >> n >> k;
  srand((int)time(0));
  for (register int i = 1; i <= n; i++) {
    val[i] = read();
    sit[val[i]] = i;
  }
  if (k == 1)
    work2();
  else if (k == n - 1)
    work3();
  else
    work1();
  return 0;
}
