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
# seishin.py
N = int(input())
*P, = map(int, input().split())
M = [0]*(N+1)
for i, p in enumerate(P):
    M[p] = i

INF = 10**9
def init(P, n):
    n0 = 2 ** (n-1).bit_length()
    data = [INF]*(n0*2)
    data[n0-1:n0+n-1] = P
    for i in range(n0-2, -1, -1):
        data[i] = min(data[2*i+1], data[2*i+2])
    return data

"""
def __get(data, a, b, k, l, r):
    if a <= l and r <= b:
        return data[k]
    if b <= l or r <= a:
        return INF
    vl = __get(data, a, b, 2*k+1, l, (l+r)//2)
    vr = __get(data, a, b, 2*k+2, (l+r)//2, r)
    return min(vl, vr)
"""

def get_data(data, l, r):
    n0 = len(data)//2
    l += n0
    r += n0
    while l < r:
        if r&1:
            r -= 1
            yield data[r-1]
        if l&1:
            yield data[l-1]
            l += 1
        l >>= 1
        r >>= 1
    yield INF

def get(data, l, r):
    #return __get(data, l, r, 0, 0, len(data)//2)
    return min(get_data(data, l, r))

d0 = init(P[0::2], N//2)
d1 = init(P[1::2], N//2)

def query_x(l, r):
    if l % 2 == 0:
        x = get(d0, l//2, r//2)
    else:
        x = get(d1, l//2, r//2)
    return x
def query_y(l, r):
    if l % 2 == 0:
        y = get(d1, l//2, r//2)
    else:
        y = get(d0, (l+1)//2, (r+1)//2)
    return y

from heapq import heappush, heappop
que = [(query_x(0, N), 0, N)]
ans = []
while que:
    x, l, r = heappop(que)
    if l+2 < r:
        xi = M[x]
        y = query_y(xi, r)

        yi = M[y]

        # [l, xi)
        if l < xi:
            heappush(que, (query_x(l, xi), l, xi))
        # [xi+1, yi)
        if xi+1 < yi:
            heappush(que, (query_x(xi+1, yi), xi+1, yi))
        # [yi+1, r)
        if yi+1 < r:
            heappush(que, (query_x(yi+1, r), yi+1, r))
    else:
        y = P[r-1]

    ans.append("%d %d" % (x, y))
print(*ans)