human
Yelisey has an array a of n integers.

If a has length strictly greater than 1, then Yelisei can apply an operation called minimum extraction to it: 

  1. First, Yelisei finds the minimal number m in the array. If there are several identical minima, Yelisey can choose any of them. 
  2. Then the selected minimal element is removed from the array. After that, m is subtracted from each remaining element. 


Thus, after each operation, the length of the array is reduced by 1.

For example, if a = [1, 6, -4, -2, -4], then the minimum element in it is a_3 = -4, which means that after this operation the array will be equal to a=[1 {- (-4)}, 6 {- (-4)}, -2 {- (-4)}, -4 {- (-4)}] = [5, 10, 2, 0].

Since Yelisey likes big numbers, he wants the numbers in the array a to be as big as possible.

Formally speaking, he wants to make the minimum of the numbers in array a to be maximal possible (i.e. he want to maximize a minimum). To do this, Yelisey can apply the minimum extraction operation to the array as many times as he wants (possibly, zero). Note that the operation cannot be applied to an array of length 1.

Help him find what maximal value can the minimal element of the array have after applying several (possibly, zero) minimum extraction operations to the array.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The next 2t lines contain descriptions of the test cases.

In the description of each test case, the first line contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the original length of the array a. The second line of the description lists n space-separated integers a_i (-10^9 ≤ a_i ≤ 10^9) — elements of the array a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

Print t lines, each of them containing the answer to the corresponding test case. The answer to the test case is a single integer — the maximal possible minimum in a, which can be obtained by several applications of the described operation to...
Using python3 can you solve the prior task?
assistant
#Here is Python code to find what maximal value can the minimal element of the array have after applying several (possibly, zero) minimum extraction operations to the array.

# Author - krishnakumar_r
import sys
import os
import math
import bisect
from collections import *
import heapq
from sys import stdin,stdout
from math import gcd,floor,sqrt,log
from collections import defaultdict as dd
from bisect import bisect_left as bl,bisect_right as br
from itertools import *
from functools import *
import random

sys.setrecursionlimit(100000000)

inp    =lambda: int(input())
strng  =lambda: input().strip()
jn     =lambda x,l: x.join(map(str,l))
strl   =lambda: list(input().strip())
mul    =lambda: map(int,input().strip().split())
mulf   =lambda: map(float,input().strip().split())
seq    =lambda: list(map(int,input().strip().split()))

ceil   =lambda x: int(x) if(x==int(x)) else int(x)+1
ceildiv=lambda x,d: x//d if(x%d==0) else x//d+1

flush  =lambda: stdout.flush()
stdstr =lambda: stdin.readline()
stdint =lambda: int(stdin.readline())
stdpr  =lambda x: stdout.write(str(x))
INF = float('inf')
MOD =1000000007

from io import BytesIO, IOBase
# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
           ...