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: Accepted)

#include <iostream>
#include <cstdio>
#include <queue>
#define N 200005
using namespace std;

struct st{
	int x, y, l, r;
} t;
int n, a[N], d[N][20];
priority_queue<st> qu;

bool operator < (st p, st q) {
	return !(a[p.x] < a[q.x]);
}

int mn(int p, int q) {
	return a[p] < a[q] ? p : q;
}

int f(int p, int q) {
	int i;
	if (p > q) return p;
	for (i = 1; 1 << i <= q - p + 1; i++);
	i--;
	return mn(d[p][i], d[q - (1 << i) + 1][i]);
}

void add(int p, int q) {
	int x, y;
	if (p > q) return;
	x = f(p, q);
	y = mn(f(x + 1, q - 1), q);
	qu.push({x, y, p, q});
}

int main()
{
	int i, j, k;
	cin >> n;
	for (i = 0; i < n; i++) scanf("%d", &a[i]), d[i][1] = i;
	for (j = 2; 1 << j <= n; j++) {
		k = 1 << j;
		for (i = 0; i + k <= n; i++) {
			d[i][j] = mn(d[i][j - 1], d[i + k / 2][j - 1]);
		}
	}
	add(0, n - 1);
	while (!qu.empty()) {
		t = qu.top(), qu.pop();
		printf("%d %d ", a[t.x], a[t.y]);
		add(t.l, t.x - 1);
		add(t.x + 1, t.y - 1);
		add(t.y + 1, t.r);
	}
    return 0;
}