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 N = 2e5 + 10 , inf = 1e9;

int n;
int num[N] , Map[N] , val[2][N*4];

void build(int x , int l , int r)
{
	if(l == r)
	{
		val[l & 1][x] = num[l];
		val[(l & 1)^1][x] = inf;
		return;
	}
	int mid = l + r >> 1;
	build(x<<1 , l  , mid);
	build(x<<1|1 , mid + 1 , r);
	val[0][x] = min(val[0][x<<1] , val[0][x<<1|1]);
	val[1][x] = min(val[1][x<<1] , val[1][x<<1|1]);
} 
void Build(){build(1 , 1 , n);}

int ask(int x , int l , int r , int L , int R , bool o)
{
	if(l >= L && r <= R)
		return val[o][x];
	int mid = l + r >> 1 , ans = inf;
	if(L <= mid) ans = min(ans , ask(x<<1 , l , mid , L , R , o));
	if(R > mid) ans = min(ans , ask(x<<1|1 , mid + 1 , r , L , R , o));
	return ans;
}
int Ask(int l , int r , bool o) {return ask(1 , 1 , n , l , r , o);}

struct Seg
{
	int l , r , val;
	bool operator< (const Seg &t) const
	{
		return val > t.val;
	}
	Seg(){}
	Seg(int l, int r) :l(l) , r(r)
	{
		val = Ask(l , r , l&1);
	}
};

priority_queue<Seg> q;

int main()
{
	scanf("%d" , &n);
	for(int i = 1 ; i <= n ; i ++)
	{
		scanf("%d" , num + i);
		Map[num[i]] = i;
	}
	Build();
	q.push(Seg(1 , n));
	while(!q.empty())
	{
		Seg now = q.top();
		q.pop();
		int l = Map[now.val];
		bool id = l & 1; 
		int r = Map[Ask(l + 1 , now.r , id^1)];
		printf("%d %d " , num[l] , num[r]);
		if(l != now.l)
			q.push(Seg(now.l , l - 1));
		if(r - l >= 2)
			q.push(Seg(l + 1 , r - 1));
		if(now.r != r)
			q.push(Seg(r + 1 , now.r));
	}
	puts("");
}