human
Write me a Python code to find \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that: given the following:
You are given a sequence a_1, a_2, ..., a_n consisting of n pairwise distinct positive integers.

Find \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that: 

  * x ≠ y.

  * x and y appear in a.

  * x~mod~y doesn't appear in a. 


Note that some x or y can belong to multiple pairs.

⌊ x ⌋ denotes the floor function — the largest integer less than or equal to x. x~mod~y denotes the remainder from dividing x by y.

If there are multiple solutions, print any of them. It can be shown that at least one solution always exists.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of testcases.

The first line of each testcase contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the length of the sequence.

The second line of each testcase contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6).

All numbers in the sequence are pairwise distinct. The sum of n over all testcases doesn't exceed 2 ⋅ 10^5.

Output

The answer for each testcase should contain \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that x ≠ y, x and y appear in a and x~mod~y doesn't appear in a. Print the pairs one after another.

You can print the pairs in any order. However, the order of numbers in the pair should be exactly such that the first number is x and the second number is y. All pairs should be pairwise distinct.

If there are multiple solutions, print any of them.

Example

Input

4
2
1 4
4
2 8 3 4
5
3 8 5 9 7
6
2 7 5 3 4 8

Output

4 1
8 2
8 4
9 5
7 5
8 7
4 3
5 2

Note

In the first testcase there are only two pairs: (1, 4) and (4, 1). \left⌊ \frac 2 2 \right⌋=1, so we have to find one pair. 1~mod~4=1, and 1 appears in a, so that pair is invalid. Thus, the only possible answer is a pair (4, 1).

In the second testcase, we chose pairs 8~mod~2=0 and 8~mod~4=0. 0 doesn't appear in a, so that answer is valid. There are multiple possible answers for that testcase.

In the third testcase, the chosen pairs are 9~mod~5=4 and 7~mod~5=2. Neither 4, nor 2, appears in a, so that...
Solve the task in PYTHON.
assistant
#Here is Python code to find \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that: 
.

import sys
testing = len(sys.argv) == 4 and sys.argv[3] == "myTest"
interactive = False
if testing:
    cmd = sys.stdout
    from time import time
    start_time = int(round(time() * 1000)) 
    readAll = open(sys.argv[1], 'r').read
    sys.stdout = open(sys.argv[2], 'w')
else:
    readAll = sys.stdin.read

# ############ ---- I/O Functions ---- ############

class InputData:
    def __init__(self):
        self.lines = readAll().split('\n')
        self.n = len(self.lines)
        self.ii = -1
    def input(self):
        self.ii += 1
        assert self.ii < self.n
        return self.lines[self.ii]

flush = sys.stdout.flush
if interactive and not testing:
    input = sys.stdin.readline
else:
    inputData = InputData()
    input = inputData.input

def intin():
    return(int(input()))
def intlin():
    return(list(map(int,input().split())))
def chrin():
    return(list(input()))
def strin():
    return input()
def lout(l, sep="\n", toStr=True):
    print(sep.join(map(str, l) if toStr else l))
def dout(*args, **kargs):
    if not testing: return
    if args: print(args[0] if len(args)==1 else args)
    if kargs: print([(k,v) for k,v in kargs.items()])
def ask(q):
    sys.stdout.write(str(q)+'\n')
    flush()
    return intin()
    
# ############ ---- I/O Functions ---- ############

# from math import ceil
# from collections import defaultdict as ddict, Counter
# from heapq import *
# from Queue import Queue

def main():
    n = intin()
    a = intlin()
    a.sort()
    m = n/2
    ans = []
    for i in xrange(m):
        ans.append('%d %d' % (a[i+1], a[0]))
    return '\n'.join(ans)

anss = []
for _ in xrange(intin()):
    anss.append(main())
    # anss.append("YES" if main() else "NO")
lout(anss)

if testing:
    sys.stdout = cmd
    print(int(round(time() * 1000))  - start_time)