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>
inline bool chkmax(T &a, T b) {
  return a < b ? (a = b, 1) : 0;
}
template <typename T>
inline bool chkmin(T &a, T b) {
  return b < a ? (a = b, 1) : 0;
}
int _, __;
char c_;
inline int read() {
  for (_ = 0, __ = 1, c_ = getchar(); !isdigit(c_); c_ = getchar())
    if (c_ == '-') __ = -1;
  for (; isdigit(c_); c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
  return _ * __;
}
inline void file() {}
const int maxn = 500010, maxm = 10000000;
int v[maxm], in[maxn], nex[maxm], beg[maxn], n, k, p[maxn], e, ans[maxn];
inline void add(int uu, int vv) {
  v[++e] = vv;
  ++in[vv];
  nex[e] = beg[uu];
  beg[uu] = e;
}
struct Item {
  int key, pos;
  bool operator<(const Item &item) const { return key < item.key; }
};
set<Item>::iterator it;
inline void init() {
  n = read();
  k = read();
  for (int i = (1), _end_ = (n); i <= _end_; ++i) p[i] = read();
  set<Item> st;
  for (int i = (1), _end_ = (n); i <= _end_; ++i) {
    if (i >= k + 1) st.erase((Item){p[i - k], i - k});
    it = st.upper_bound((Item){p[i], i});
    if (it != st.end()) add(i, (*it).pos);
    st.insert((Item){p[i], i});
  }
  st.clear();
  for (int i = (1), _end_ = (k - 1); i <= _end_; ++i)
    st.insert((Item){p[i], i});
  for (int i = (1), _end_ = (n); i <= _end_; ++i) {
    st.erase((Item){p[i], i});
    st.insert((Item){p[i + k - 1], i + k - 1});
    it = st.upper_bound((Item){p[i], i});
    if (it != st.end()) add(i, (*it).pos);
    st.erase((Item){p[i - k + 1], i - k + 1});
  }
}
inline void toposort() {
  priority_queue<int, vector<int>, greater<int> > q;
  for (int i = (1), _end_ = (n); i <= _end_; ++i)
    if (!in[i]) q.push(i);
  int now = 0;
  while (!q.empty()) {
    int u = q.top();
    q.pop();
    ans[u] = ++now;
    for (int i = beg[u]; i; i = nex[i]) {
      --in[v[i]];
      if (!in[v[i]]) q.push(v[i]);
    }
  }
}
inline void print() {
  for (int i = (1), _end_ = (n); i <= _end_; ++i) printf("%d\n", ans[i]);
}
int main() {
  file();
  init();
  toposort();
  print();
  return 0;
}
