Hard to Play

MightyHorse is playing a music game called osu!.

 

After playing for several months, MightyHorse discovered the way of calculating score in osu!:

1. While playing osu!, player need to click some circles following the rhythm. Each time a player clicks, it will have three different points: 300, 100 and 50, deciding by how clicking timing fits the music.

2. Calculating the score is quite simple. Each time player clicks and gets P points, the total score will add P, which should be calculated according to following formula:

 

P = Point * (Combo * 2 + 1)

 

Here Point is the point the player gets (300, 100 or 50) and Combo is the number of consecutive circles the player gets points previously - That means if the player doesn't miss any circle and clicks the i^th circle, Combo should be i - 1.

Recently MightyHorse meets a high-end osu! player. After watching his replay, MightyHorse finds that the game is very hard to play. But he is more interested in another problem: What's the maximum and minimum total score a player can get if he only knows the number of 300, 100 and 50 points the player gets in one play?

As the high-end player plays so well, we can assume that he won't miss any circle while playing osu!; Thus he can get at least 50 point for a circle.

Input

There are multiple test cases.

The first line of input is an integer T (1 ≤ T ≤ 100), indicating the number of test cases.

For each test case, there is only one line contains three integers: A (0 ≤ A ≤ 500) - the number of 300 point he gets, B (0 ≤ B ≤ 500) - the number of 100 point he gets and C (0 ≤ C ≤ 500) - the number of 50 point he gets.

Output

For each test case, output a line contains two integers, describing the minimum and maximum total score the player can get.

Sample Input

1
2 1 1 

Sample Output

2050 3950

 

#include <stdio.h>

int main()
{
    int t,a,b,c,s1,s2;
    scanf("%d",&t);
    while(t--)
    {
        s1=s2=0;
        scanf("%d %d %d",&a,&b,&c);
        for(int i=0; i<a; i++) s1+=300*(2*i+1);
        for(int i=0; i<b; i++) s1+=100*(2*(a+i)+1);
        for(int i=0; i<c; i++) s1+=50*(2*(a+b+i)+1);

        for(int i=0; i<c; i++) s2+=50*(2*i+1);
        for(int i=0; i<b; i++) s2+=100*(2*(c+i)+1);
        for(int i=0; i<a; i++) s2+=300*(2*(c+b+i)+1);

        printf("%d %d\n",s1,s2);
    }

    return 0;
}