1208_A. XORinacci
Cengiz recently learned Fibonacci numbers and now he is studying different algorithms to find them. After getting bored of reading them, he came with his own new type of numbers that he named XORinacci numbers. He defined them as follows: 

  * f(0) = a; 
  * f(1) = b; 
  * f(n) = f(n-1) ⊕ f(n-2) when n > 1, where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR). 



You are given three integers a, b, and n, calculate f(n).

You have to answer for T independent test cases.

Input

The input contains one or more independent test cases.

The first line of input contains a single integer T (1 ≤ T ≤ 10^3), the number of test cases.

Each of the T following lines contains three space-separated integers a, b, and n (0 ≤ a, b, n ≤ 10^9) respectively.

Output

For each test case, output f(n).

Example

Input


3
3 4 2
4 5 0
325 265 1231232


Output


7
4
76

Note

In the first example, f(2) = f(0) ⊕ f(1) = 3 ⊕ 4 = 7.
```python
import sys
from collections import defaultdict as dd
from collections import deque
from functools import *
from fractions import Fraction as f
from copy import *
from bisect import *	
from heapq import *
from math import *
from itertools import permutations ,product
 
def eprint(*args):
    print(*args, file=sys.stderr)
zz=1
 
#sys.setrecursionlimit(10**6)
if zz:
	input=sys.stdin.readline
else:	
	sys.stdin=open('input.txt', 'r')
	sys.stdout=open('all.txt','w')
def inc(d,c):
	d[c]=d[c]+1 if c in d else 1
def bo(i):
	return ord(i)-ord('A')	
def li():
	return [int(xx) for xx in input().split()]
def fli():
	return [float(x) for x in input().split()]	
def comp(a,b):
	if(a>b):
		return 2
	return 2 if a==b else 0		
def gi():	
	return [xx for xx in input().split()]
def fi():
	return int(input())
def pro(a): 
	return reduce(lambda a,b:a*b,a)		
def swap(a,i,j): 
	a[i],a[j]=a[j],a[i]	
def si():
	return list(input().rstrip())	
def mi():
	return 	map(int,input().split())			
def gh():
	sys.stdout.flush()
def isvalid(i,j):
	return 0<=i<n and 0<=j<n	
def bo(i):
	return ord(i)-ord('a')	
def graph(n,m):
	for i in range(m):
		x,y=mi()
		a[x].append(y)
		a[y].append(x)


t=fi()


while t>0:
	t-=1
	a,b,n=mi()
	n+=1
	if n%3==0:
		print(a^b)
	elif n%3==1:
		print(a)
	else:
		print(b)		

	

```