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;
int n,a[200005],w[2][19][200005];
struct O{int x,y;};
vector <O> f; vector <int> e[100005];
struct cmp{
	bool operator()(int a,int b){
		return f[a].x>f[b].x;
	}
};
priority_queue <int,vector<int>,cmp> d;
int Min(int x,int y){return a[x]<a[y]?x:y;}
int get(int o,int l,int r){
	int i=log2(r-l+1);
	return Min(w[o][i][r],w[o][i][l-1+(1<<i)]);
}
int dfs(int o,int l,int r){
	if (l>r) return 0;
	int p=get(o,l,r),q=get(o^1,p+1,r),x=f.size(),y;
	f.push_back({a[p],a[q]});
	y=dfs(o,l,p-1); if (y) e[x].push_back(y);
	y=dfs(o^1,p+1,q-1); if (y) e[x].push_back(y);
	y=dfs(o,q+1,r); if (y) e[x].push_back(y);
	return x;
}
int main(){
	scanf("%d",&n); a[0]=n+3;
	for (int i=1;i<=n;++i) scanf("%d",&a[i]);
	for (int i=1;i<=n;++i)
		w[0][0][i]=i&1?0:i,w[1][0][i]=i&1?i:0;
	for (int k=0;k<2;++k)
	for (int i=1;i<18;++i)
	for (int j=1<<i;j<=n;++j)
	w[k][i][j]=Min(w[k][i-1][j-(1<<i-1)],w[k][i-1][j]);
	d.push(dfs(1,1,n));
	while (!d.empty()){
		n=d.top(); d.pop();
		printf("%d %d ",f[n].x,f[n].y);
		for (int i=0;i<e[n].size();++i) d.push(e[n][i]);
	}
	puts("");
	return 0;
}