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 <cstdio>
#include <set>
#include <queue>
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
const int N = 1e6+10;
int n,k,a[N],b[N],deg[N];
vector<int> g[N];
set<int> L,R;
priority_queue<int> q;
int main() {
	scanf("%d%d",&n,&k);
	REP(i,1,n) scanf("%d",a+i),b[a[i]]=i;
	REP(i,1,k) R.insert(a[i]);
	REP(i,1,n) {
		auto p = L.upper_bound(a[i]);
		if (p!=L.end()) g[b[*p]].push_back(i),++deg[i];
		p = R.upper_bound(a[i]);
		if (p!=R.end()) g[b[*p]].push_back(i),++deg[i];
		L.insert(a[i]),R.erase(a[i]);
		if (i-k+1>0) L.erase(a[i-k+1]);
		if (i+k<=n) R.insert(a[i+k]);
	}
	REP(i,1,n) if (!deg[i]) q.push(i);
	PER(i,1,n) {
		int u = q.top(); q.pop();
		a[u] = i;
		for (int v:g[u]) if (!--deg[v]) q.push(v);
	}
	REP(i,1,n) printf("%d\n",a[i]);
}
