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...
#Mamm
assistant
#Here is Python code to find \left⌊ \frac n 2 \right⌋ different pairs of integers x and y such that: 
.
a don't raises quitter.................................................
#manan is improving, keep hustling manan bhai ;)
from collections import deque as de
import math
import sys
from math import sqrt as sq
from math import floor as fl
from math import ceil as ce
from sys import stdin, stdout
import time

sys.setrecursionlimit(10**7)
from collections import Counter as cnt
from functools import  reduce

from itertools import groupby as gb
#from fractions import Fraction as fr
from bisect import bisect_left as bl, bisect_right as br
from typing import Coroutine

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
    
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def __iter__(self):
        return iter(self.data)
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0
    def len(self):
        return len(self.data)
    

arrStack = My_stack() 

class graph():
    def __init__(self):
        self.graphdict = {}
    def add_edge(self, node, neighbor):
        if node not in self.graphdict:
            self.graphdict[node]=[neighbor]
        else:
            self.graphdict[node].append(neighbor)
    def show_edge(self):
        for node in self.graphdict:
            for neighbors in self.graphdict:
                print(node, neighbors)
    def show_graph(self):
        return self.graphdict

    

    
            

#decimal to binary   
def decimalToBinary(n): 
    return bin(n).replace("0b", "")
#binary to decimal
def binarytodecimal(n):
    return int(n,2)

def isPrime(n) : 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True
	if (n % 2 == 0 or n % 3 == 0) : 
		return False
	i = 5
	while(i * i <= n) : 
		if...