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;
typedef long long ll;
const ll inf=0x3f3f3f3f;
const int N=2e5+20;
int mn[2][N][30],n,a[2][N],pos[N];
void init(int v)
{
	for(int i=0;i<n;i++)
		mn[v][i][0]=a[v][i];
	for(int j=1;(1<<j)<=n;j++)
		for(int i=0;(i+(1<<j)-1)<n;i++)
			mn[v][i][j]=min(mn[v][i][j-1],mn[v][i+(1<<(j-1))][j-1]);
}
int rmq(int l,int r,int v)
{
	int k=0;
	while((1<<(k+1))<=r-l+1)
		k++;
	return min(mn[v][l][k],mn[v][r-(1<<k)+1][k]);
}
struct node{
	int x,l,r;
	bool operator < (const node &t)const{
		return x>t.x;
	}
};
int main()
{
	int x;
	while(cin>>n)
	{
		for(int i=0;i<n;i++)
		{
			scanf("%d",&x);
			pos[x]=i;
			a[i&1][i]=x;
			a[(i&1)^1][i]=inf;
		}
		init(0),init(1);
		priority_queue<node> q;
		x=rmq(0,n-1,0);
		q.push({x,0,n-1});
		while(!q.empty())
		{
			node x=q.top();
			q.pop();
			int v1=x.x,v2=rmq(pos[x.x]+1,x.r,(pos[x.x]&1)^1);
			printf("%d %d ",v1,v2);
			v1=pos[v1],v2=pos[v2];
			if(x.l!=v1)
				q.push({rmq(x.l,v1-1,x.l&1),x.l,v1-1});
			if(v1+1!=v2)
				q.push({rmq(v1+1,v2-1,(v1+1)&1),v1+1,v2-1});
			if(v2!=x.r)
				q.push({rmq(v2+1,x.r,(v2+1)&1),v2+1,x.r});
		}
		printf("\n");
	}
	return 0;
 
} 