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 Main {
	static Scanner sc = new Scanner(System.in);
	public static void main(String args[]){
		
		int time = sc.nextInt();
		while(time>0){
			long x = sc.nextLong();
			long y = sc.nextLong();
			long a = sc.nextLong();
			long b = sc.nextLong();
			if(x==0||y==0){
				if(x==0){
					System.out.println(a*y);
				}else if(y==0){
					System.out.println(a*x);
				}else{
					System.out.println(0);
				}
			}else{
				long max = Math.max(x, y);
				long min = Math.min(x, y);
				if(a*2>b){
					System.out.println(a*(max-min)+b*min);
				}else{
					System.out.println((max+min)*a);
				}
			}
			
			time--;
		}
	}
	
}
