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 maxn=2e5+10;

int n,p[maxn];

struct SegTree{
	int op,pos[maxn<<2];
#define lc (nd<<1)
#define rc (nd<<1|1)
#define mid ((s+t)>>1)

	void init(int _op) {op=_op;}

	void build(int nd,int s,int t)
	{
		if (s==t)
		{
			pos[nd]=(s&1)==op?s:0;
			return;
		}
		build(lc,s,mid);
		build(rc,mid+1,t);
		pos[nd]=p[pos[lc]]<p[pos[rc]]?pos[lc]:pos[rc];
	}

	int query(int nd,int s,int t,int l,int r)
	{
		if (l<=s&&t<=r) return pos[nd];
		int Ans=0,tmp;
		if (l<=mid) tmp=query(lc,s,mid,l,r),Ans=p[Ans]<p[tmp]?Ans:tmp;
		if (r> mid) tmp=query(rc,mid+1,t,l,r),Ans=p[Ans]<p[tmp]?Ans:tmp;
		return Ans;
	}
#undef lc
#undef rc
#undef mid
}S[2];

struct Interval{
	int l,r,u,v;
	Interval() {}
	Interval(int l,int r):l(l),r(r)
	{
		u=S[l&1].query(1,1,n,l,r);
		v=S[(l&1)^1].query(1,1,n,u+1,r);
	}
	bool operator < (const Interval &i) const {return p[u]>p[i.u];}
};

priority_queue<Interval> q;

int main()
{
#ifdef h10
	freopen("E.in","r",stdin);
	freopen("E.out","w",stdout);
#endif
	int i;
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&p[i]);
	p[0]=0x7fffffff;
	S[0].init(0); S[0].build(1,1,n);
	S[1].init(1); S[1].build(1,1,n);
	q.push(Interval(1,n));
	while (!q.empty())
	{
		Interval t=q.top(); q.pop();
		printf("%d %d ",p[t.u],p[t.v]);
		if (t.l<t.u-1)
			q.push(Interval(t.l,t.u-1));
		if (t.u+1<t.v-1)
			q.push(Interval(t.u+1,t.v-1));
		if (t.v+1<t.r)
			q.push(Interval(t.v+1,t.r));
	}
}
