Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

4
3 2 4 1

Sample Output 1

3 1 2 4

The solution above is obtained as follows:

p

q

(3, 2, 4, 1)

()

↓

↓

(3, 1)

(2, 4)

↓

↓

()

(3, 1, 2, 4)

Sample Input 2

2
1 2

Sample Output 2

1 2

Sample Input 3

8
4 6 3 2 8 5 7 1

Sample Output 3

3 1 2 7 4 6 8 5

The solution above is obtained as follows:

p

q

(4, 6, 3, 2, 8, 5, 7, 1)

()

↓

↓

(4, 6, 3, 2, 7, 1)

(8, 5)

↓

↓

(3, 2, 7, 1)

(4, 6, 8, 5)

↓

↓

(3, 1)

(2, 7, 4, 6, 8, 5)

↓

↓

()

(3, 1, 2, 7, 4, 6, 8, 5)

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <iostream>
#include <queue>
#include <string>
#include <algorithm>
#include <cstdio>
#include <stack>
#include <vector>
#include <map>
#include <ctime>
#include <cmath>
#include <string.h>
#define inf 0x7fffffff
using namespace std;
struct I
{
    int x,y;
};


int a,b,c,k,n,m,i,e,d,o[211010],l[200212],dx[20]={1,0,-1,0},dy[20]={0,1,0,-1},dz[20]={1,-1};
long long x,y,z[110];
//map <string,int> m;
int j[1060];

stack<int> s;
//char r[55];
//deque<int> d;
struct P
{
    int x,y,z;
    bool operator<(const P &a)const
    {

        return z<a.z;
    }
};
double u;
//P u[500000];
vector <int> v[211010];
priority_queue<P> q;
string r,r1[100];
map<string,int > p;
//A==65
//Z==90

//a==97

void f()
{
    for(;q.size()&&(l[q.top().x]==0||l[q.top().y]==0);q.pop());
    if(q.size()==0) return;
    P u=q.top();
    q.pop();
    l[u.x]=0;
    l[u.y]=0;
    if(u.x>1&&u.y<a)
    q.push({u.x-1,u.y+1,o[u.x-1]+o[u.x+1]});
    f();
    printf("%d %d ",o[u.x],o[u.y]);
}

int main()
{
    scanf("%d",&a);
    for(int t=1;t<=a;l[t]=1,t++)
        scanf("%d",&o[t]);
    for(int t=2;t<=a;t++)
    {
        q.push({t-1,t,o[t]+o[t-1]});
    }
    f();
}