H - Hard to Play

 

MightyHorse is playing a music game called osu!.

 

After playing for several months, MightyHorsediscovered 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, MightyHorsefinds 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

12 1 1

Sample Output

2050 3950

1 #include
      
        2 #include
       
         3 #include
        
          4 int main() 5 { 6     int t; 7     int a,b,c; 8     int p1,p2,p3,p4,p5,p6; 9     int mins,maxs;10     while(scanf("%d",&t)!=EOF)11     {12         mins=0;13         maxs=0;14         while(t--)15         {16            /* mins=0;17             maxs=0;*/18             p1=p2=p3=p4=p5=p6=0;19             scanf("%d%d%d",&a,&b,&c);20             for(int i=0;i