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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#define MAXN 200010
#define X first
#define Y second

typedef long long ll;

int N;
int a[MAXN], f[MAXN][19], pos[MAXN];
int lg[MAXN], bin[MAXN];

vector<pair<int, int> > ans;

int GetMin(int l, int r)
{
	int len = (r - l + 2) / 2;
	int k = lg[len];
	if ((r - l + 1) & 1)
		return pos[min(f[l][k], f[r - 2 * (bin[k] - 1)][k])];
	else return pos[min(f[l][k], f[r - 2 * (bin[k] - 1) - 1][k])]; 
}

vector<pair<int, int> > Solve(int l, int r)
{
	if (l > r) 
	{
		vector<pair<int, int> > res;
		return res;
	}
	int ps1 = GetMin(l, r);
	int ps2 = GetMin(ps1 + 1, r);
	
	vector<pair<int, int> > lis[3], res;
	lis[0] = Solve(l, ps1 - 1);;
	lis[1] = Solve(ps1 + 1, ps2 - 1);
	lis[2] = Solve(ps2 + 1, r);
	int p[3] = {0};
	
	res.push_back(make_pair(a[ps1], a[ps2]));
	while (p[0] < lis[0].size() || p[1] < lis[1].size() || p[2] < lis[2].size())
	{
		int k = -1; pair<int, int> Min;
		for (int i = 0; i <= 2; ++i)
		{
			if (p[i] == lis[i].size()) continue;
			if (k == -1 || lis[i][p[i]] < Min)
				k = i, Min = lis[i][p[i]];
		}
		p[k]++; res.push_back(Min);
	}
	return res;
}

int main()
{
	//freopen("in.in", "r", stdin);
	//freopen("out.out", "w", stdout);
	
	for (int i = 0; i <= 18; ++i)
		bin[i] = 1 << i, lg[bin[i]] = i;
	for (int i = 2; i < MAXN; ++i)
		if (!lg[i]) lg[i] = lg[i - 1];
	
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i)
		scanf("%d", &a[i]), f[i][0] = a[i], pos[a[i]] = i;
	
	for (int j = 1, h = 2; h <= N; ++j, h <<= 1)
		for (int i = 1; i + 2 * (h - 1) <= N; ++i)
			f[i][j] = min(f[i][j - 1], f[i + h][j - 1]);
	ans = Solve(1, N);
	
	for (int i = 0; i < ans.size(); ++i)
	{
		if (i) putchar(' ');
		printf("%d %d", ans[i].first, ans[i].second);
	}
	printf("\n");
	return 0;
}