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<cstdio>
#include<queue>
using namespace std;
inline int read()
{
	int x;char c;
	while((c=getchar())<'0'||c>'9');
	for(x=c-'0';(c=getchar())>='0'&&c<='9';)x=x*10+c-'0';
	return x;
}
#define N 262144
#define INF 0x3FFFFFFF
#define mp make_pair
#define tp pq.top()
#define F first
#define S second
int t[N*2+5],t1[N*2+5],t2[N*2+5],p[N];
priority_queue<pair<pair<int,int>,pair<int,int> > > pq;
void build(int*t){for(int i=N;--i;)t[i]=min(t[i<<1],t[i<<1|1]);}
int query(int*t,int l,int r)
{
	int res=INF;
	for(l+=N-1,r+=N+1;l^r^1;l>>=1,r>>=1)
	{
		if(~l&1)res=min(res,t[l+1]);
		if( r&1)res=min(res,t[r-1]);
	}
	return res;
}
void push(int l,int r)
{
	if(l>r)return;
	int x=query(l&1?t1:t2,l,r);
	pq.push(mp(mp(-x,query(l&1?t2:t1,p[x]+1,r)),mp(l,r)));
}
int main()
{
	int n=read(),i,x,y,l,r;
	for(i=1;i<=n;++i)p[x=read()]=i,t[i+N]=x,t1[i+N]=i&1?x:INF,t2[i+N]=i&1?INF:x;
	build(t);build(t1);build(t2);
	for(push(1,n);!pq.empty();)
	{
		x=-tp.F.F;y=tp.F.S;l=tp.S.F;r=tp.S.S;
		printf("%d %d ",x,y);pq.pop();
		push(l,p[x]-1);push(p[x]+1,p[y]-1);push(p[y]+1,r);
	}
}