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 <cstdio>
#include <queue>
using namespace std;
struct st{int l, r, x, y;};
int n, a[200005], m[200005][20];
priority_queue<st> pq;
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 l, int r) {
	int j=0;
	while(1<<(j+1) <= r-l+1) j++;
	return mn(m[l][j], m[r-(1<<j)+1][j]);
}
 
void add(int l, int r) {
	int x, y;
	if(l>r) return;
	x = f(l, r);
	y = x+1==r ? r : mn(f(x+1, r-1), r);
	pq.push({l,r,x,y});
}
 
int main() {
	int i, j, k;
	scanf("%d", &n);
	for(i=0; i<n; i++) scanf("%d", a+i), m[i][1] = i;
	for(j=2; ; j++) {
		k = 1<<j;
		if(k>n) break;
		for(i=0; i+k<=n; i++) m[i][j] = mn(m[i][j-1], m[i+k/2][j-1]);
	}
	for(add(0, n-1); !pq.empty(); ) {
		st t = pq.top(); pq.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;
}