There are two problems in a contest.
Problem A is worth 500 points at the start of the contest.
Problem B is worth 1000 points at the start of the contest.

Once the contest starts, after each minute:
Maximum points of Problem A reduce by 2 points .
Maximum points of Problem B reduce by 4 points.

It is known that Chef requires X minutes to solve Problem A correctly and Y minutes to solve Problem B correctly.

Find the maximum number of points Chef can score if he optimally decides the order of attempting both the problems.

------ Input Format ------

• First line will contain T, number of test cases. Then the test cases follow.
• Each test case contains of a single line of input, two integers X and Y - the time required to solve problems A and B in minutes respectively.

------ Output Format ------

For each test case, output in a single line, the maximum number of points Chef can score if he optimally decides the order of attempting both the problems.

------ Constraints ------

$1 ≤ T ≤ 1000$
$1 ≤ X, Y ≤ 100$

----- Sample Input 1 ------
4
10 20
8 40
15 15
20 10

----- Sample Output 1 ------
1360
1292
1380
1400

----- explanation 1 ------
Test Case $1$: If Chef attempts in the order $A \rightarrow B$ then he submits Problem A after $10$ minutes and Problem B after $30$ minutes.
Thus, he gets $500 - 10 \cdot 2 = 480$ points for problem A and $1000 - 30 \cdot 4 = 880$ points for problem B. Thus, total $480+880=1360$ points for both the problems.

If Chef attempts in the order $B \rightarrow A$ then he submits Problem B after $20$ minutes and Problem A after $30$ minutes.
Thus, he gets $1000 - 20 \cdot 4 = 920$ points for Problem B and $500 - 30 \cdot 2 = 440$ points for Problem A. Thus total $920+440=1360$ points for both the problems.

So, in both cases Chef gets $1360$ points in total.

Test Case $2$: If Chef attempts in the order $A \rightarrow B$ then he submits Problem A after $8$ minutes and Problem B after $48$ minutes.
Thus, he gets $500 - 8 \cdot 2 = 484$ points for problem A and $1000 - 48 \cdot 4 = 808$ points for problem B. Thus, total $484+808=1292$ points for both the problems.

If Chef attempts in the order $B \rightarrow A$ then he submits Problem B after $40$ minutes and Problem A after $48$ minutes.
Thus, he gets $1000 - 40 \cdot 4 = 840$ points for Problem B and $500 - 48 \cdot 2 = 404$ points for Problem A. Thus total $840+404=1244$ points for both the problems.

So, Chef will attempt in the order $A \rightarrow B$ and thus obtain $1292$ points.

Test Case $3$: If Chef attempts in the order $A \rightarrow B$ then he submits Problem A after $15$ minutes and Problem B after $30$ minutes.
Thus, he gets $500 - 15 \cdot 2 = 470$ points for problem A and $1000 - 30 \cdot 4 = 880$ points for problem B. Thus, total $470+880=1350$ points for both the problems.

If Chef attempts in the order $B \rightarrow A$ then he submits Problem B after $15$ minutes and Problem A after $30$ minutes.
Thus, he gets $1000 - 15 \cdot 4 = 940$ points for Problem B and $500 - 30 \cdot 2 = 440$ points for Problem A. Thus total $940+440=1380$ points for both the problems.

So, Chef will attempt in the order $B \rightarrow A$ and thus obtain $1380$ points.

Test Case $4$: If Chef attempts in the order $A \rightarrow B$ then he submits Problem A after $20$ minutes and Problem B after $30$ minutes.
Thus, he gets $500 - 20 \cdot 2 = 460$ points for problem A and $1000 - 30 \cdot 4 = 880$ points for problem B. Thus, total $460+880=1340$ points for both the problems.

If Chef attempts in the order $B \rightarrow A$ then he submits Problem B after $10$ minutes and Problem A after $30$ minutes.
Thus, he gets $1000 - 10 \cdot 4 = 960$ points for Problem B and $500 - 30 \cdot 2 = 440$ points for Problem A. Thus total $960+440=1400$ points for both the problems.

So, Chef will attempt in the order $B \rightarrow A$ and thus obtain $1400$ points.

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