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;
inline int read() {
  int x = 0, ch;
  while (!isdigit(ch = getchar()))
    ;
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  return x;
}
template <typename T>
inline void chkmin(T& x, T y) {
  x = x < y ? x : y;
}
const int MAXN = 500000;
int a[MAXN + 1];
int d[MAXN + 1];
int to[MAXN + 1][2];
int T[MAXN << 2];
priority_queue<int, vector<int>, greater<int> > q;
inline void Modify(int nd, int l, int r, int p, int v) {
  T[nd] = v;
  if (l == r) return;
  int mid = (l + r) >> 1;
  if (p <= mid)
    Modify((nd << 1), l, mid, p, v);
  else
    Modify((nd << 1 | 1), mid + 1, r, p, v);
}
inline int Query(int nd, int l, int r, int L, int R) {
  if (L <= l && r <= R) return T[nd];
  int mid = (l + r) >> 1, res = 2147483647;
  if (L <= mid) chkmin(res, Query((nd << 1), l, mid, L, R));
  if (mid < R) chkmin(res, Query((nd << 1 | 1), mid + 1, r, L, R));
  return res;
}
int main() {
  int n = read(), k = read();
  for (int i = 1; i <= n; i++) a[read()] = i;
  memset(T, 63, sizeof(T));
  for (int i = n; i >= 1; i--) {
    int L = Query(1, 1, n, max(1, a[i] - k + 1), a[i]);
    int R = Query(1, 1, n, a[i], min(n, a[i] + k - 1));
    if (L <= n && a[i] - a[L] < k) d[a[L]]++, to[a[i]][0] = a[L];
    if (R <= n && a[R] - a[i] < k) d[a[R]]++, to[a[i]][1] = a[R];
    Modify(1, 1, n, a[i], i);
  }
  int x, u;
  for (int i = 1; i <= n; i++)
    if (!d[i]) q.push(i);
  while (!q.empty()) {
    printf("%d\n", x = q.top());
    q.pop();
    u = to[x][0];
    if (u) {
      d[u]--;
      if (!d[u]) q.push(u);
    }
    u = to[x][1];
    if (u) {
      d[u]--;
      if (!d[u]) q.push(u);
    }
  }
  return 0;
}
