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;
template <typename T>
bool chkmax(T &a, T b) {
  return (a < b) ? a = b, 1 : 0;
}
template <typename T>
bool chkmin(T &a, T b) {
  return (a > b) ? a = b, 1 : 0;
}
inline int read() {
  int x = 0, fh = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') fh = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + (ch ^ '0');
  return x * fh;
}
vector<int> G[500005];
int tree[2000005];
int p[500005];
int q[500005];
int in[500005];
priority_queue<int, vector<int>, greater<int> > Q;
void update(int rt, int l, int r, int pos, int val) {
  if (l == r) {
    tree[rt] = val;
    return;
  }
  if (pos <= ((l + r) >> 1))
    update(rt << 1, l, ((l + r) >> 1), pos, val);
  else
    update(rt << 1 | 1, ((l + r) >> 1) + 1, r, pos, val);
  tree[rt] = min(tree[rt << 1], tree[rt << 1 | 1]);
}
int now;
int query(int rt, int l, int r, int L, int R) {
  if (L > R) return 0x3f3f3f3f;
  if (L <= l && r <= R) return tree[rt];
  int res = 0x3f3f3f3f;
  if (L <= ((l + r) >> 1)) chkmin(res, query(rt << 1, l, ((l + r) >> 1), L, R));
  if (R > ((l + r) >> 1))
    chkmin(res, query(rt << 1 | 1, ((l + r) >> 1) + 1, r, L, R));
  return res;
}
int n, k;
int ans[500005], top;
int main() {
  memset(tree, 0x3f3f3f3f, sizeof(tree));
  n = read();
  k = read();
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i) {
    p[i] = read();
    q[p[i]] = i;
  }
  for (int i = (n), _end_ = (int)(1); i >= _end_; --i) {
    int now = query(1, 1, n, q[i] + 1, min(q[i] + k - 1, n));
    if (now != 0x3f3f3f3f) {
      G[q[i]].push_back(q[now]);
      in[q[now]]++;
    }
    now = query(1, 1, n, max(1, q[i] - k + 1), q[i] - 1);
    if (now != 0x3f3f3f3f) {
      G[q[i]].push_back(q[now]);
      in[q[now]]++;
    }
    update(1, 1, n, q[i], i);
  }
  for (int i = (1), _end_ = (int)(n); i <= _end_; ++i)
    if (!in[i]) Q.push(i);
  while (!Q.empty()) {
    int x = Q.top();
    Q.pop();
    ans[++top] = x;
    for (int v : G[x]) {
      if (--in[v] == 0) Q.push(v);
    }
  }
  for (int i = (1), _end_ = (int)(top); i <= _end_; ++i) printf("%d\n", ans[i]);
  return 0;
}
