Problem: You have n sticks of the given lengths.

Your task is to choose exactly four of them in such a way that they can form a rectangle. No sticks can be cut to pieces, each side of the rectangle must be formed by a single stick. No stick can be chosen multiple times. It is guaranteed that it is always possible to choose such sticks.

Let S be the area of the rectangle and P be the perimeter of the rectangle. 

The chosen rectangle should have the value (P^2)/(S) minimal possible. The value is taken without any rounding.

If there are multiple answers, print any of them.

Each testcase contains several lists of sticks, for each of them you are required to solve the problem separately.

Input

The first line contains a single integer T (T ≥ 1) — the number of lists of sticks in the testcase.

Then 2T lines follow — lines (2i - 1) and 2i of them describe the i-th list. The first line of the pair contains a single integer n (4 ≤ n ≤ 10^6) — the number of sticks in the i-th list. The second line of the pair contains n integers a_1, a_2, ..., a_n (1 ≤ a_j ≤ 10^4) — lengths of the sticks in the i-th list.

It is guaranteed that for each list there exists a way to choose four sticks so that they form a rectangle.

The total number of sticks in all T lists doesn't exceed 10^6 in each testcase.

Output

Print T lines. The i-th line should contain the answer to the i-th list of the input. That is the lengths of the four sticks you choose from the i-th list, so that they form a rectangle and the value (P^2)/(S) of this rectangle is minimal possible. You can print these four lengths in arbitrary order.

If there are multiple answers, print any of them.

Example

Input

3
4
7 2 2 7
8
2 8 1 4 8 2 1 5
5
5 5 5 5 5


Output

2 7 7 2
2 2 1 1
5 5 5 5

Note

There is only one way to choose four sticks in the first list, they form a rectangle with sides 2 and 7, its area is 2 ⋅ 7 = 14, perimeter is 2(2 + 7) = 18. (18^2)/(14) ≈ 23.143.

The second list contains subsets of four sticks that can form rectangles with sides (1, 2), (2, 8) and (1, 8). Their values are (6^2)/(2) = 18, (20^2)/(16) = 25 and (18^2)/(8) = 40.5, respectively. The minimal one of them is the rectangle (1, 2).

You can choose any four of the 5 given sticks from the third list, they will form a square with side 5, which is still a rectangle with sides (5, 5).

Solution (in PYTHON):

Solution:
#!/usr/bin/env python
"""
This file is part of https://github.com/Cheran-Senthil/PyRival.

Copyright 2018 Cheran Senthilkumar all rights reserved,
Cheran Senthilkumar <hello@cheran.io>
Permission to use, modify, and distribute this software is given under the
terms of the MIT License.

"""
from __future__ import division, print_function

import cmath
import itertools
import math
import operator as op
# import random
import sys
from atexit import register
from bisect import bisect_left, bisect_right
from collections import Counter
# from copy import deepcopy
# from decimal import Decimal
# from difflib import SequenceMatcher
# from fractions import Fraction
# from heapq import heappop, heappush

if sys.version_info[0] < 3:
    # from cPickle import dumps
    from io import BytesIO as stream
    # from Queue import PriorityQueue, Queue
else:
    # from functools import reduce
    from io import StringIO as stream
    from math import gcd
    # from pickle import dumps
    # from queue import PriorityQueue, Queue


if sys.version_info[0] < 3:
    class dict(dict):
        """dict() -> new empty dictionary"""
        def items(self):
            """D.items() -> a set-like object providing a view on D's items"""
            return dict.iteritems(self)

        def keys(self):
            """D.keys() -> a set-like object providing a view on D's keys"""
            return dict.iterkeys(self)

        def values(self):
            """D.values() -> an object providing a view on D's values"""
            return dict.itervalues(self)

    def gcd(x, y):
        """gcd(x, y) -> int
        greatest common divisor of x and y
        """
        while y:
            x, y = y, x % y
        return x

    input = raw_input
    range = xrange

    filter = itertools.ifilter
    map = itertools.imap
    zip = itertools.izip


def sync_with_stdio(sync=True):
    """Set whether the standard Python streams are allowed to buffer their I/O.

    Args:
        sync (bool, optional): The new synchronization setting.

    """
    global input, flush

    if sync:
        flush = sys.stdout.flush
    else:
        sys.stdin = stream(sys.stdin.read())
        input = lambda: sys.stdin.readline().rstrip('\r\n')

        sys.stdout = stream()
        register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))


def main():
    t = int(input())
    for i in range(t):
        n = int(input())
        a = Counter(map(int, input().split(' '))).items()

        found = False
        sides = []

        for ai in a:
            if ai[1] > 3:
                print(ai[0], ai[0], ai[0], ai[0])
                found = True
                break
            elif ai[1] > 1:
                sides.append(ai[0])

        if found:
            continue

        sides.sort()
        pre = sides.pop()
        m = 10001

        for j in range(len(sides)):
            nex = sides.pop()

            if (pre / nex) + (nex / pre) < m:
                m = (pre / nex) + (nex / pre)
                ans = [pre, pre, nex, nex]

            pre = nex

        print(*ans)


if __name__ == '__main__':
    sync_with_stdio(False)
    main()