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 F() {
  char ch;
  int x, a;
  while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-')
    ;
  if (ch == '-')
    ch = getchar(), a = -1;
  else
    a = 1;
  x = ch - '0';
  while (ch = getchar(), ch >= '0' && ch <= '9')
    x = (x << 1) + (x << 3) + ch - '0';
  return a * x;
}
const int N = 5e5 + 10;
int n, k, p[N], q[N];
int t[N << 2], cnt;
int get(int x, int y) {
  if (!x) return y;
  if (!y) return x;
  return min(x, y);
}
void modify(int x, int v) {
  t[x += cnt] = v;
  for (x >>= 1; x; x >>= 1) t[x] = get(t[x << 1], t[x << 1 | 1]);
}
int query(int l, int r) {
  int ans = 0;
  l = max(l, 1);
  r = min(r, n);
  if (l > r) return 0;
  for (l += cnt - 1, r += cnt + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
    if (~l & 1) ans = get(ans, t[l ^ 1]);
    if (r & 1) ans = get(ans, t[r ^ 1]);
  }
  return ans;
}
vector<int> edge[N];
int deg[N];
priority_queue<int> pq;
int main() {
  (((n) = F()), ((k) = F()));
  for (int i = 1; i <= n; ++i) ((p[i]) = F()), q[p[i]] = i;
  for (cnt = 1; cnt < n + 2; cnt <<= 1)
    ;
  --cnt;
  for (int i = n; i; --i) {
    int x = q[query(q[i] - k + 1, q[i] - 1)];
    if (x) edge[q[i]].push_back(x), ++deg[x];
    x = q[query(q[i] + 1, q[i] + k - 1)];
    if (x) edge[q[i]].push_back(x), ++deg[x];
    modify(q[i], i);
  }
  for (int i = 1; i <= n; ++i)
    if (!deg[i]) pq.push(-i);
  while (!pq.empty()) {
    int u = -pq.top();
    pq.pop();
    printf("%d\n", u);
    for (auto v : edge[u]) {
      --deg[v];
      if (!deg[v]) pq.push(-v);
    }
  }
  return 0;
}
