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
    using namespace std;
     
    int A[SIZE];
    int pos[SIZE];
    int back[SIZE];
    int n,K;
    const int SIZE=500005;
     
    int main()
    {
    	scanf("%d %d",&n,&K);
    	for(int i=0;i<n;i++)
    	{
    		scanf("%d",&A[i]);
    		A[i]--;
    		pos[A[i]]=i;
    	}
    	memset(back,-1,sizeof(back));
    	set <int> st;
    	st.insert(n);
    	set <int>::iterator it,its;
    	for(int i=n-1;i>=0;i--)
    	{
    		it=st.lower_bound(pos[i]-K+1);
    		int v=*it;
    		back[pos[i]]=back[v];
    		back[v]=pos[i];
    		while(it!=st.end())
    		{
    			int f=*it;
    			if(f>pos[i]) break;
    			its=it;it++;
    			st.erase(its);
    		}
    		st.insert(pos[i]);
    	}
    	stack <int> ans;
    	int now=n;
    	for(int i=n-1;i>=0;i--)
    	{
    		now=back[now];
    		pos[i]=now;
    	}
    	for(int i=0;i<n;i++) A[pos[i]]=i;
    	for(int i=0;i<n;i++) printf("%d\n",A[i]+1);
    	return 0;
    }