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
// copied
#include<cstdio>
#include<queue>
using namespace std;
inline int read() {
	int x;
	char c;
	while ((c = getchar()) < '0' || c > '9')
		;
	for (x = c - '0'; (c = getchar()) >= '0' && c <= '9';)
		x = x * 10 + c - '0';
	return x;
}
#define N 262144
#define INF 0x3FFFFFFF
#define mp make_pair
#define tp pq.top()
#define F first
#define S second
int t1[N * 2 + 5], t2[N * 2 + 5], p[N];
priority_queue<pair<pair<int, int>, pair<int, int> > > pq;
void build(int*t) {
	for (int i = N; --i;)
		t[i] = min(t[i << 1], t[i << 1 | 1]);
}
int query(int*t, int l, int r) {
	int res = INF;
	for (l += N - 1, r += N + 1; l ^ r ^ 1; l >>= 1, r >>= 1) {
		if (~l & 1)
			res = min(res, t[l + 1]);
		if (r & 1)
			res = min(res, t[r - 1]);
	}
	return res;
}
void push(int l, int r) {
	if (l > r)
		return;
	int x = query(l & 1 ? t1 : t2, l, r);
	pq.push(mp(mp(-x, query(l & 1 ? t2 : t1, p[x] + 1, r)), mp(l, r)));
}
int main() {
	int n = read(), i, x, y, l, r;
	for (i = 1; i <= n; ++i)
		p[x = read()] = i,  t1[i + N] = i & 1 ? x : INF, t2[i + N] =
				i & 1 ? INF : x;
	build(t1);
	build(t2);
	for (push(1, n); !pq.empty();) {
		x = -tp.F.F;
		y = tp.F.S;
		l = tp.S.F;
		r = tp.S.S;
		printf("%d %d ", x, y);
		pq.pop();
		push(l, p[x] - 1);
		push(p[x] + 1, p[y] - 1);
		push(p[y] + 1, r);
	}
}
