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.
def min(a,b):
	if(a>=b):
		return b
	return a



t = int(input())


while t > 0:
	
	x,y = map(int, input().split())
	a,b = map(int, input().split())
	op1 = 100000000000000000000000000000000000
	op2 = 1000000000000000000000000000000000000
	op3 = 0
	if((x<0 and y < 0) or (x > 0 and y > 0)):		
		if(x<0):
			x = x*-1
		if(y<0):
			y = y*-1
		op1 = x-y
		if(op1 < 0):
			op1*=-1
		op1*=a;
		op2 = min(x,y)*b
	if(x<0):
		x = x*-1
	if(y<0):
		y = y*-1
	
	op3 = x*a + y*a
	r = min(op1+op2,op3)
	
	print(r)
	t-=1
	
