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;
double EPS = 1e-9;
int INF = 1e9 + 7;
;
long long INFLL = 1e17;
double pi = acos(-1);
int dirx[8] = {-1, 0, 0, 1, -1, -1, 1, 1};
int diry[8] = {0, 1, -1, 0, -1, 1, -1, 1};
clock_t first_attempt = clock();
inline void cek_time() {
  clock_t cur = clock() - first_attempt;
  cerr << "TIME : " << (double)cur / CLOCKS_PER_SEC << endl;
}
inline void OPEN(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const int maxn = 5e5 + 5;
int n, K;
int pos[maxn], ans[maxn], ret[maxn];
int main() {
  scanf("%d%d", &n, &K);
  assert(n <= 8000);
  for (int k = (int)1; k < (int)n + 1; k++) {
    int a;
    scanf("%d", &a);
    pos[a] = k;
  }
  for (int k = (int)n - 1; k >= (int)1; k--) {
    for (int i = (int)1; i < (int)n + 1; i++) printf("%d ", pos[i]);
    printf("\n");
    for (int i = (int)k; i < (int)n; i++) {
      if (pos[i] < pos[i + 1] + K) break;
      swap(pos[i], pos[i + 1]);
    }
  }
  for (int k = (int)1; k < (int)n + 1; k++) ans[pos[k]] = k;
  for (int k = (int)1; k < (int)n + 1; k++) printf("%d\n", ans[k]);
  return 0;
}
