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.util.*;
public class Road {
	public static void main(String[] args) {
		Scanner sc= new Scanner(System.in);
		int cases = sc.nextInt();
		long[][] a = new long[cases][4];
		for (int i=0;i<cases;i++) {
			a[i][0]=sc.nextInt();
			a[i][1]=sc.nextInt();
			a[i][2]=sc.nextInt();
			a[i][3]=sc.nextInt();
			
		}
		for (int i=0;i<cases;i++) {
			if (a[i][3]>2*a[i][2]) {System.out.println((a[i][0]+a[i][1])*a[i][2]);}
			else {
				long s=a[i][1];
				long f=a[i][0];
				if (a[i][1]<a[i][0]) {s=a[i][0];f=a[i][1];}
				System.out.println((f)*a[i][3] + (s-f)*a[i][2]);	
			}
			
		}
				
	}

}
