You must have tried to solve the Rubik’s cube. You might even have succeeded at it. Rubik’s cube is a 3x3x3 cube which has 6 different color for each face.The Rubik’s cube is made from 26 smaller pieces which are called cubies. There are 6 cubies at the centre of each face and these comprise of a single color. There are 8 cubies at the 8 corners which comprise of exactly 3 colors. The 12 reamaining cubies comprise of exactly 2 colors.

Apple has come up with a variation of the Rubik’s Cube, it’s the Rubik’s cuboid which has different colors on its 6 faces. The Rubik’s Cuboid comes in various sizes represented by M x N x O (M,N,O are natural numbers). Apple is giving away 100 Rubik’s cuboid for free to people who can answer a simple questions. Apple wants to know, in a Rubik’s cuboid with arbitrary dimensions, how many cubies would be there, which comprise of exactly 2 color.

-----Input-----
The input contains several test cases.The first line of the input contains an integer T denoting the number of test cases.

Each test case comprises of 3 natural numbers, M,N & O, which denote the dimensions of the Rubiks Cuboid.

-----Output-----
For each test case you are required to output the number of  cubies which comprise of 2 squares, each of which is of a different color.

-----Constraints-----
- 1 ≤ T ≤ <1000
- 1 ≤ M ≤ <100000
- 1 ≤ N ≤ <100000
- 1 ≤ O ≤ <100000

-----Example-----
Input:
1
3
3
3

Output:

12

Difficulty: interview

Starter code:

Solutions:
for _ in range(int(input())):
 m=int(input())
 n=int(input())
 o=int(input())
 ans=4*(m+n+o)-24
 if(ans <= 0):
  print('0')
 else:
  print(ans)


import sys

for __ in range(eval(input())) :
 m , n , o = eval(input()) , eval(input()) , eval(input())
 if m==1 and n==1 and o==1 :
  print(0)
  continue
 print((m-2)*4+(n-2)*4+(o-2)*4)


for tt in range(eval(input())):
 m=eval(input())
 n=eval(input())
 o=eval(input())
 if m==1 and n==1 and o==1:
  print('0')
  continue
 if m!=1 and n!=1 and o!=1:
  ans = abs(4*(m-2)+4*(n-2)+4*(o-2))
  print(ans)
 elif (m==1 and n>1 and o>1):
  ans = (n*o) - ((2*n)+2*(o-2))
  print(ans)
 elif (n==1 and m>1 and o>1):
  ans = (m*o) - ((2*m)+2*(o-2))
  print(ans)
 elif (o==1 and m>1 and n>1):
  ans = (m*n) - ((2*m)+2*(n-2))
  print(ans)


from sys import stdin, stdout
t = int(stdin.readline())
while t:
 t -= 1
 m = int(stdin.readline())
 n = int(stdin.readline())
 o = int(stdin.readline())
 if m>2 and n>2 and o>2:
  ans = 4*(m+n+o-6)
 elif m==1 or n==1 or o==1 or(m==2 and n==2 and o==2):
  ans=0
 else:
  ans = (m-2) if m > 2 else 0
  ans += (n-2) if m > 2 else 0
  ans += (o-2) if m > 2 else 0
 stdout.write(str(ans)+'\n')

for t in range(int(input())):
 m=int(input())
 n=int(input())
 o=int(input())
 if m==1 and (n==1 and o==1):
  print(0)
 else:
  print((m+n+o-6)*4)

for _ in range(int(input())):
 a=int(input())
 b=int(input())
 c=int(input())
 if a==1 or b==1 or c==1:
  print(0) 
 else:
  print(4*(a+b+c-6))