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;
const int INF = 0x3f3f3f3f;
int n, k, a[N], pos[N], deg[N];
set<int> L, R;
vector<int> G[N];
priority_queue<int> q;
void toposort() {
  for (int i = 1; i <= n; i++)
    if (deg[i] == 0) q.push(-i);
  while (!q.empty()) {
    int x = -q.top();
    q.pop();
    printf("%d\n", x);
    for (int i = 0; i < G[x].size(); i++) {
      int y = G[x][i];
      if (--deg[y] == 0) q.push(-y);
    }
  }
}
int Min[N << 2];
void build(int rt, int l, int r) {
  Min[rt] = INF;
  if (l == r) return;
  int mid = l + r >> 1;
  build(rt << 1, l, mid);
  build(rt << 1 | 1, mid + 1, r);
}
void update(int rt, int l, int r, int q, int v) {
  if (l == r) {
    Min[rt] = min(Min[rt], v);
    return;
  }
  int mid = l + r >> 1;
  if (q <= mid) update(rt << 1, l, mid, q, v);
  if (q > mid) update(rt << 1 | 1, mid + 1, r, q, v);
  Min[rt] = min(Min[rt << 1], Min[rt << 1 | 1]);
}
int query(int rt, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return Min[rt];
  int mid = l + r >> 1;
  int ret = INF;
  if (ql <= mid) ret = min(ret, query(rt << 1, l, mid, ql, qr));
  if (qr > mid) ret = min(ret, query(rt << 1 | 1, mid + 1, r, ql, qr));
  return ret;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pos[a[i]] = i;
  build(1, 1, n);
  for (int i = n; i; i--) {
    int t1 = query(1, 1, n, max(1, pos[i] - k + 1), pos[i]);
    if (t1 <= n) G[pos[i]].push_back(pos[t1]), deg[pos[t1]]++;
    int t2 = query(1, 1, n, pos[i], min(n, pos[i] + k - 1));
    if (t2 <= n) G[pos[i]].push_back(pos[t2]), deg[pos[t2]]++;
    update(1, 1, n, pos[i], i);
  }
  toposort();
  return 0;
}
