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 <iostream>
#include <cstdio>
#include <queue>
using namespace std;
struct st{int l, r, x, v;};
int n, bs, lt, rt, fr, to, a[200005], tr[2][1<<18];
priority_queue<st> pq;
bool operator < (st p, st q) {return p.v>q.v;}

int f(int p, int q, int r) {
	int t, mn=1e9, mi;
	p = bs + p/2;
	q = bs + q/2;
	while(p<q) {
		if(p%2==1) {
			t = tr[r][p];
			if(a[t]<mn) mn=a[t], mi=t;
			p++;
		}
		if(q%2==0) {
			t = tr[r][q];
			if(a[t]<mn) mn=a[t], mi=t;
			q--;
		}
		p/=2, q/=2;
	}
	if(p==q) {
		t = tr[r][p];
		if(a[t]<mn) mn=a[t], mi=t;
	}
	return mi;
}

int main() {
	int i, k;
	scanf("%d", &n);
	for(i=0; i<n; i++) scanf("%d", a+i);
	for(bs=1; bs<n/2; bs*=2);
	for(i=0; i<n/2; i++) tr[0][bs+i] = i*2, tr[1][bs+i] = i*2+1;
	for(i=bs-1; i; i--) {
		lt = tr[0][i*2], rt = tr[0][i*2+1];
		tr[0][i] = a[lt] < a[rt] ? lt : rt;
		lt = tr[1][i*2], rt = tr[1][i*2+1];
		tr[1][i] = a[lt] < a[rt] ? lt : rt;
	}
	k = f(0, n-2, 0);
	pq.push({0, n-1, k, a[k]});
	while(!pq.empty()) {
		st t = pq.top(); pq.pop();
		fr=t.l, to=t.r, lt=t.x;
		rt = f(lt+1, to, (lt+1)%2);
		printf("%d %d ", t.v, a[rt]);
		if(fr<lt) {
			k = f(fr, lt-2, fr%2);
			pq.push({fr, lt-1, k, a[k]});
		}
		if(rt-lt>1) {
			k = f(lt+1, rt-2, (lt+1)%2);
			pq.push({lt+1, rt-1, k, a[k]});
		}
		if(rt<to) {
			k = f(rt+1, to-1, (rt+1)%2);
			pq.push({rt+1, to, k, a[k]});
		}
	}
	return 0;
}