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>
#define ull unsigned long long
#define pq priority_queue
#define ll long long
#define pp pair<ull,ull>
#define rush() int T;scanf("%d",&T);while(T--)  
#define forr(n) for(int i=1;i<=n;i++)
using namespace std;
const int maxn = 200005;  
const ll mod = 1e9+7;  
const int INF = 0x3f3f3f;  
const double eps = 1e-9;  
int a[2][maxn]; 
int n; 
struct node
{
	int v,l,r;
	bool operator <(const node&b)const
	{
		return v>b.v;
	}
}no[maxn];
int s[2][maxn][18];
int pos[maxn];
void initRmq(int t)
{
	forr(n)
	{
		s[t][i][0]=a[t][i];
	}
	for(int j=1;(1<<j)<=n;j++)
	for(int i=1;i+(1<<j)-1<=n;i++)
	s[t][i][j]=min(s[t][i][j-1],s[t][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(s[v][l][k],s[v][r-(1<<k)+1][k]);
}
int main()
{
	node tmp;
	int temp;
	pq <node>q;
	cin>>n;
	forr(n)
	{
		scanf("%d",&temp);
		pos[temp]=i;
		a[i&1][i]=temp;
		a[(i&1)^1][i]=INF;
	}
	initRmq(1);
	initRmq(0);
	tmp.v=RMQ(1,n,1);
	tmp.l=1;
	tmp.r=n;
	q.push(tmp);
	int v1,v2;
	while(!q.empty())
	{
		node x=q.top();
		q.pop();
		v1=x.v,v2=RMQ(pos[v1]+1,x.r,(pos[v1]&1)^1);
		printf("%d %d ",v1,v2);
		v1 = pos[v1], v2 = pos[v2];
		if(v1!=x.l)
		{
			tmp.v=RMQ(x.l,v1-1,x.l&1);
			tmp.l=x.l;
			tmp.r=v1-1;
			q.push(tmp);
		}
		if(v1!=v2-1)
		{
			tmp.v=RMQ(v1+1,v2-1,(v1+1)&1);
			tmp.l=v1+1;
			tmp.r=v2-1;
			q.push(tmp);			
		}
		if(v2!=x.r)
		{
			tmp.v=RMQ(v2+1,x.r,(v2+1)&1);
			tmp.l=v2+1;
			tmp.r=x.r;
			q.push(tmp);			
		}
	}
	return 0;
}