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
from heapq import heappop, heappush
n = int(input())
p = [int(x) for x in input().split()]


class SegmentTree:  # 0-indexed
    def __init__(self, array, operation=min, identity=10**30):
        self.identity = identity
        self.n = len(array)
        self.N = 1 << (self.n - 1).bit_length()
        self.tree = [self.identity] * 2 * self.N
        self.opr = operation
        for i in range(self.n):
            self.tree[i+self.N-1] = array[i]
        for i in range(self.N-2, -1, -1):
            self.tree[i] = self.opr(self.tree[2*i+1], self.tree[2*i+2])

    def values(self):
        return self.tree[self.N-1:]

    def update(self, k, x):
        k += self.N-1
        self.tree[k] = x
        while k+1:
            k = (k-1)//2
            self.tree[k] = self.opr(self.tree[k*2+1], self.tree[k*2+2])

    def query(self, p, q):  # [p,q)
        if q <= p:
            print("Oops!  That was no valid number.  Try again...")
            exit()
        p += self.N-1
        q += self.N-2
        res = self.identity
        while q-p > 1:
            if p & 1 == 0:
                res = self.opr(res, self.tree[p])
            if q & 1 == 1:
                res = self.opr(res, self.tree[q])
                q -= 1
            p = p//2
            q = (q-1)//2
        if p == q:
            res = self.opr(res, self.tree[p])
        else:
            res = self.opr(self.opr(res, self.tree[p]), self.tree[q])
        return res


ind = [0]*(n+1)
Odd = SegmentTree([10**30]*n)
Even = SegmentTree([10**30]*n)

for i in range(n):
    ind[p[i]] = i
    if i % 2 == 0:
        Even.update(i, p[i])
    else:
        Odd.update(i, p[i])

cand = []
heappush(cand, (Even.query(0, n), 0, n, True))

q = []
for _ in range(n//2):
    first, l, r, is_even = heappop(cand)
    if is_even:
        second = Odd.query(ind[first]+1, r)
        q.extend([first, second])
        if l < ind[first]:
            heappush(cand, (Even.query(l, ind[first]), l, ind[first], True))
        if ind[first] + 1 < ind[second]:
            heappush(
                cand, (Odd.query(ind[first]+1, ind[second]), ind[first]+1, ind[second], False))
        if ind[second]+1 < r:
            heappush(
                cand, (Even.query(ind[second], r), ind[second]+1, r, True))
    else:
        second = Even.query(ind[first]+1, r)
        q.extend([first, second])
        if l < ind[first]:
            heappush(cand, (Odd.query(l, ind[first]), l, ind[first], False))
        if ind[first] + 1 < ind[second]:
            heappush(
                cand, (Even.query(ind[first]+1, ind[second]), ind[first]+1, ind[second], True))
        if ind[second]+1 < r:
            heappush(
                cand, (Odd.query(ind[second], r), ind[second]+1, r, False))

print(*q)
