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;
const int N = 5e5 + 10;
struct dabt {
  int next, to;
} e[N];
int head[N], k, n, m, tree[N << 2], ru[N], pos[N], p[N], cnt, ans[N];
priority_queue<int, vector<int>, greater<int> > pn;
inline void add(int from, int to) {
  e[++k] = (dabt){head[from], to};
  head[from] = k;
  ru[to]++;
}
inline int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline int mi(int a, int b) { return a < b ? a : b; }
inline int mx(int a, int b) { return a > b ? a : b; }
int sum(int l, int r, int st, int ed, int o) {
  if (st <= l && r <= ed) return tree[o];
  if (l == r) return 2e9;
  int mid = l + (r - l) / 2, ans = 2e9;
  if (mid >= st) ans = mi(ans, sum(l, mid, st, ed, o << 1));
  if (mid < ed) ans = mi(ans, sum(mid + 1, r, st, ed, o << 1 | 1));
  return ans;
}
void upd(int l, int r, int pos, int kk, int o) {
  if (l == r) {
    tree[o] = kk;
    return;
  }
  int mid = l + (r - l) / 2;
  if (pos <= mid)
    upd(l, mid, pos, kk, o << 1);
  else
    upd(mid + 1, r, pos, kk, o << 1 | 1);
  tree[o] = mi(tree[o << 1], tree[o << 1 | 1]);
}
inline void topo() {
  for (int i = 1; i <= n; i++)
    if (!ru[i]) pn.push(i);
  while (!pn.empty()) {
    int x = pn.top();
    pn.pop();
    ans[x] = ++cnt;
    for (int i = head[x]; i; i = e[i].next) {
      int v = e[i].to;
      if (--ru[v] == 0) pn.push(v);
    }
  }
}
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= (n << 2); i++) tree[i] = 2e9;
  for (int i = 1; i <= n; i++) {
    pos[p[i] = read()] = i;
  }
  for (int i = n, x; i; i--) {
    x = sum(1, n, pos[i] + 1, mi(pos[i] + m - 1, n), 1);
    if (x < 2e9) add(pos[i], pos[x]);
    x = sum(1, n, mx(pos[i] - m + 1, 1), pos[i] - 1, 1);
    if (x < 2e9) add(pos[i], pos[x]);
    upd(1, n, pos[i], i, 1);
  }
  topo();
  for (int i = 1; i <= n; i++) printf("%d\n", ans[i]);
  return 0;
}
