You are given two integers x and y. You can perform two types of operations: 

  1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation: 
    * x = 0, y = 6; 
    * x = 0, y = 8; 
    * x = -1, y = 7; 
    * x = 1, y = 7. 

  2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation: 
    * x = -1, y = 6; 
    * x = 1, y = 8. 



Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.

Calculate the minimum amount of dollars you have to spend on it.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.

The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).

The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).

Output

For each test case print one integer — the minimum amount of dollars you have to spend.

Example

Input


2
1 3
391 555
0 0
9 4


Output


1337
0

Note

In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.

In the second test case both integers are equal to zero initially, so you dont' have to spend money.
import java.io.*;
import java.util.Arrays;
import java.util.Scanner;
 
public class Main {
 
 
 
    public static void main(String[] args)  {
 
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
            int t = in.nextInt();
            for (int testCase = 1; testCase <= t; ++testCase) {
                int x = in.nextInt();
                int y = in.nextInt();
                int a = in.nextInt();
                int b = in.nextInt();
 
                long cost=0;
                long min = Math.min(x,y);
 
                if((2*a)<b)
                {
                    //use two as instead of b
                    cost+=(long)((2*a)*min);
 
                }
                else{
                    //use bs first
                    cost+=(long)(b*min);
                }
                x-=min;
                y-=min;
 
                // one of x or y is zero
                long remaining = Math.abs(x-y);
                cost+=(long)(a*remaining);
 
                System.out.println(cost);
            }
 
 
    }
}