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
// KALAM
# include<bits/stdc++.h>

using namespace std;

const int N = 500000 + 77;
int n , k , cur = 1 , a[N] , P[N] , Mx[N << 2] , A[N];
bool M[N];
vector < int > adj[N];
void Update(int x , int y , int l = 1 , int r = n + 1 , int id = 1) {
   if(r - l < 2) {
      Mx[id] = y;
      return ;
   }
   int mid = ((l + r) >> 1);
   if(x < mid)
      Update(x , y , l , mid , id << 1);
   else
      Update(x , y , mid , r , id << 1 ^ 1);
   Mx[id] = max(Mx[id << 1] , Mx[id << 1 ^ 1]);
}
int Get(int ql , int qr , int l = 1 , int r = n + 1 , int id = 1) {
   if(qr <= l || r <= ql)
      return 0;
   if(ql <= l && r <= qr)
      return Mx[id];
   int mid = ((l + r) >> 1);
   return max(Get(ql , qr , l , mid , id << 1) , Get(ql , qr , mid , r , id << 1 ^ 1));
}
void dfs(int v) {
   M[v] = 1;
   for(int u : adj[v])
      if(! M[u])
         dfs(u);
   A[cur ++] = v;
}
int main() {
   scanf("%d %d" , & n , & k);
   for(int i = 1;i <= n;++ i)
      scanf("%d" , a + i) , P[a[i]] = i;
   for(int i = 1;i <= n;++ i) {

      int le = Get(max(P[i] - k + 1 , 1) , P[i]);
      int ri = Get(P[i] , min(P[i] + k , n + 1));
      if(le)
         adj[P[i]].push_back(P[le]);
      if(ri)
         adj[P[i]].push_back(P[ri]);
      Update(P[i] , i);
   }
   for(int i = 1;i <= n;++ i)
      if(! M[i])
         dfs(i);
   for(int i = 1;i <= n;++ i)
      a[A[i]] = i;
   for(int i = 1;i <= n;++ i)
      printf("%d\n" , a[i]);
   return 0;
}
