Let N be a positive even number.

We have a permutation of (1, 2, ..., N), p = (p_1, p_2, ..., p_N). Snuke is constructing another permutation of (1, 2, ..., N), q, following the procedure below.

First, let q be an empty sequence. Then, perform the following operation until p becomes empty:

* Select two adjacent elements in p, and call them x and y in order. Remove x and y from p (reducing the length of p by 2), and insert x and y, preserving the original order, at the beginning of q.



When p becomes empty, q will be a permutation of (1, 2, ..., N).

Find the lexicographically smallest permutation that can be obtained as q.

Constraints

* N is an even number.
* 2 ≤ N ≤ 2 × 10^5
* p is a permutation of (1, 2, ..., N).

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the lexicographically smallest permutation, with spaces in between.

Examples

Input

4
3 2 4 1


Output

3 1 2 4


Input

2
1 2


Output

1 2


Input

8
4 6 3 2 8 5 7 1


Output

3 1 2 7 4 6 8 5
#include<bits/stdc++.h>

using namespace std;

const int MX=200011;

int n,p[MX],ip[MX];
int lg2[MX];
int mn[19][MX];
struct Q{
	int l,r,mn;
	Q(int _l=0,int _r=0,int _mn=0):l(_l),r(_r),mn(_mn){}
	inline bool operator<(const Q &x)const{return mn>x.mn;}
};

inline int q(int l,int r){
	int c=((r-l)>>1)+1;
	int L=lg2[c];
	return min(mn[L][l],mn[L][l+((c-(1<<L))<<1)]);
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n;
	for(int i=1;i<=n;i++)cin>>p[i];
	for(int i=1;i<=n;i++)ip[p[i]]=i;
	lg2[1]=0;
	for(int i=2;i<=n;i++)lg2[i]=lg2[i>>1]+1;
	for(int i=1;i<=n;i++)mn[0][i]=p[i];
	for(int i=1;(1<<i)<=n;i++){
		for(int j=1;j+(1<<(i+1))-2<=n;j+=2)mn[i][j]=min(mn[i-1][j],mn[i-1][j+(1<<i)]);
		for(int j=2;j+(1<<(i+1))-2<=n;j+=2)mn[i][j]=min(mn[i-1][j],mn[i-1][j+(1<<i)]);
	}
	static priority_queue<Q>qi;
	qi.push(Q(1,n,q(1,n)));
	for(int i=1;i<=(n>>1);i++){
		Q r=qi.top();qi.pop();
		cout<<r.mn<<' ';
		int x=ip[r.mn],y=q(ip[r.mn]+1,r.r);
		cout<<y<<' ';
		y=ip[y];
		if(r.l<x)qi.push(Q(r.l,x-1,q(r.l,x-1)));
		if(x+1<y)qi.push(Q(x+1,y-1,q(x+1,y-1)));
		if(y<r.r)qi.push(Q(y+1,r.r,q(y+1,r.r)));
	}
	cout<<endl;
	return 0;
}
