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>
//#include<iostream>
using namespace std;
int n,a[210000],ans[210000],Log2[210000],ODD[21][210000],EVEN[21][210000];
struct node{
	int x,y,l,r;
	bool operator<(const node &b) const{ return a[x]>a[b.x];}
} u;
priority_queue<node> que;
void init(){
	int m=(n>>1);
	for (int i=1;i<=m;i++){
		ODD[0][i]=(i<<1)-1;
		EVEN[0][i]=(i<<1);
	}
	for (int j=1;(1<<j)<=m;j++)
		for (int i=1;i+(1<<j)-1<=m;i++){
			if (a[ODD[j-1][i]]<a[ODD[j-1][i+(1<<(j-1))]]) ODD[j][i]=ODD[j-1][i];
			else ODD[j][i]=ODD[j-1][i+(1<<(j-1))];
			if (a[EVEN[j-1][i]]<a[EVEN[j-1][i+(1<<(j-1))]]) EVEN[j][i]=EVEN[j-1][i];
			else EVEN[j][i]=EVEN[j-1][i+(1<<(j-1))];
		}
}
int query(int (*now)[210000],int l,int r){
	int k=Log2[r-l+1];
	if (a[now[k][l]]<a[now[k][r-(1<<k)+1]]) return now[k][l];
	else return now[k][r-(1<<k)+1];
}
int getmin(int l,int r,bool x){
	if (x) return query(ODD,(l+1)>>1,(r+1)>>1);
	else return query(EVEN,l>>1,r>>1);
}
void pushin(int l,int r){
	if (l>r) return;
	int t1=getmin(l,r,l&1),t2=getmin(t1+1,r,r&1);
	que.push((node){t1,t2,l,r}); //cout << t1 << ' ' << t2 << ' ' << l << ' ' << r << endl;
}
int main(){
	Log2[0]=-1;
	for (int i=1;i<=200000;i++) Log2[i]=Log2[i>>1]+1;
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	init(); pushin(1,n);
	int cnt=0;
	while (!que.empty()){
		u=que.top(); que.pop();
		ans[++cnt]=a[u.x]; ans[++cnt]=a[u.y];
		pushin(u.l,u.x-1); pushin(u.x+1,u.y-1); pushin(u.y+1,u.r);		
	}
	for (int i=1;i<cnt;i++) printf("%d ",ans[i]);
	printf("%d\n",ans[cnt]);
	return 0;
}
