for _ in range(int(input())):
b, c, d = map(int, input().split())
a = 0
f = 0
for i in range(62):
ch = 1 << i
if d & ch:
if ch ^ (b & ch):
a |= ch
f |= ch & c
elif b & ch:
if ch & c:
a |= ch
else:
f |= True
print([a, -1][f!=0])
Explanation
For each of the 62 bits i.e $[2^0, 2^{61}]$
if $d_{i}$ is set;
and $b_{i}$ is not set;
then $a_{i}$ must be set;
raise a flag if $c_{i}$ is set, as it would unset this bit in the final result.
else if $b_{i}$ is set; save us the stress and do nothing.
else if $d_{i}$ is not set and $b_{i}$ is set i.e we don't need this bit but $b_{i}$ gives it anyway;
then we need $c_{i}$ to be set to take it out, and if we must get $c_{i}$ then we have to set $a_{i}$;
else if $c_{i}$ is not set, we raise a flag as we have an unwanted set bit that we can't take out.
Finally if the flag was raised print $-1$ else print $a$.
Shorter code for C.
Explanation
For each of the 62 bits i.e $[2^0, 2^{61}]$
Finally if the flag was raised print $-1$ else print $a$.