Compare the triplet

Alice and Bob each created one problem for HackerRank. A reviewer rates the two challenges, awarding points on a scale from  to  for three categories: problem clarity, originality, and difficulty.

We define the rating for Alice's challenge to be the triplet , and the rating for Bob's challenge to be the triplet .

Your task is to find their comparison points by comparing  with ,  with , and  with .

• If , then Alice is awarded  point.
• If , then Bob is awarded  point.
• If , then neither person receives a point.

Comparison points is the total points a person earned.

Given  and , can you compare the two challenges and print their respective comparison points?

Input Format

The first line contains  space-separated integers, , , and , describing the respective values in triplet . 
The second line contains  space-separated integers, , , and , describing the respective values in triplet .

Constraints

Output Format

Print two space-separated integers denoting the respective comparison points earned by Alice and Bob.

Sample Input

5 6 7
3 6 10

Sample Output

1 1 

Explanation

In this example:

Now, let's compare each individual score:

• , so Alice receives  point.
• , so nobody receives a point.
• , so Bob receives  point.

Alice's comparison score is , and Bob's comparison score is . Thus, we print 1 1 (Alice's comparison score followed by Bob's comparison score) on a single line.

Solution:

#!/bin/python3

import os
import sys

#
# Complete the solve function below.
#
def solve(a0, a1, a2, b0, b1, b2):
    #
    # Write your code here.
    #
    a = (1 if a0 > b0 else 0) + (1 if a1 > b1 else 0) + (1 if a2 > b2 else 0)
    b = (1 if a0 < b0 else 0) + (1 if a1 < b1 else 0) + (1 if a2 < b2 else 0)
    return (a,b)
        

if __name__ == '__main__':
    f = open(os.environ['OUTPUT_PATH'], 'w')

    a0A1A2 = input().split()

    a0 = int(a0A1A2[0])

    a1 = int(a0A1A2[1])

    a2 = int(a0A1A2[2])

    b0B1B2 = input().split()

    b0 = int(b0B1B2[0])

    b1 = int(b0B1B2[1])

    b2 = int(b0B1B2[2])

    result = solve(a0, a1, a2, b0, b1, b2)

    f.write(' '.join(map(str, result)))
    f.write('\n')

    f.close()