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.


/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
import java.io.BufferedInputStream;
import java.util.Arrays;

import java.util.Scanner;

/**
 *
 * @author Elizabeth
 */
public class Main {

    public static void main(String[] args) {
        Scanner leer = new Scanner(new BufferedInputStream(System.in));
        int tc = leer.nextInt();
        while (tc-- > 0) {
            long x = leer.nextLong();
            long y = leer.nextLong();
            long a = leer.nextLong();
            long b = leer.nextLong();
            if (x == y && x == 0) {
                System.out.println("0");
            } else if (x == y) {

                long minD=Math.min((x*a+y*a),(x*b));
                System.out.println(minD);
            } else {
                long max = Math.max(x, y);
                long min = Math.min(x, y);
                long dolares = (max - min) * a;
                long res =  Math.min((min * b),(min*2*a));
               // res=Math.min(dolares, (dolares+(2*min*a)));
                System.out.println(res+dolares);
            }
        }
    }

}
