There is an array $a$ with $n-1$ integers. Let $x$ be the bitwise XOR of all elements of the array. The number $x$ is added to the end of the array $a$ (now it has length $n$), and then the elements are shuffled.

You are given the newly formed array $a$. What is $x$? If there are multiple possible values of $x$, you can output any of them.


-----Input-----

The input consists of multiple test cases. The first line contains an integer $t$ ($1 \leq t \leq 1000$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer $n$ ($2 \leq n \leq 100$) — the number of integers in the resulting array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($0 \le a_i \le 127$) — the elements of the newly formed array $a$.

Additional constraint on the input: the array $a$ is made by the process described in the statement; that is, some value of $x$ exists.


-----Output-----

For each test case, output a single integer — the value of $x$, as described in the statement. If there are multiple possible values of $x$, output any of them.


-----Examples-----

Input
4
4
4 3 2 5
5
6 1 10 7 10
6
6 6 6 6 6 6
3
100 100 0
Output
3
7
6
0


-----Note-----

In the first test case, one possible array $a$ is $a=[2, 5, 4]$. Then $x = 2 \oplus 5 \oplus 4 = 3$ ($\oplus$ denotes the bitwise XOR), so the new array is $[2, 5, 4, 3]$. Afterwards, the array is shuffled to form $[4, 3, 2, 5]$.

In the second test case, one possible array $a$ is $a=[1, 10, 6, 10]$. Then $x = 1 \oplus 10 \oplus 6 \oplus 10 = 7$, so the new array is $[1, 10, 6, 10, 7]$. Afterwards, the array is shuffled to form $[6, 1, 10, 7, 10]$.

In the third test case, all elements of the array are equal to $6$, so $x=6$.

In the fourth test case, one possible array $a$ is $a=[100, 100]$. Then $x = 100 \oplus 100 = 0$, so the new array is $[100, 100, 0]$. Afterwards, the array is shuffled to form $[100, 100, 0]$. (Note that after the shuffle, the array can remain the same.)

Source: codeforces | Difficulty: EASY
Tags: Bit manipulation

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Solutions:
t = int(input())
for _ in range(t):
	n = int(input())
	l = list(map(int, input().split()))
	print(l[-1])


for _ in range(int(input())):
	a = int(input())
	sList = list(map(int, input().split()))
	print(sList[-1])


t = int(input())
r = []
for _ in range(t):
	n = int(input())
	a = list(map(int, input().split()))
	r.append(a[-1])
for i in r:
	print(i)