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.
#alpha = [ chr(i) for i in range(ord('a'),ord('z')+1)]
arr  = [[],[]]
def out1(x,n,m):
    # print(n,m)
    # if arr[n][m] != True:
    #     return False
    if x <= 0:

        return True
    if n <= 0 :
        if (x-10*m) > 0 :
            # arr[n][m] = False
            return False
        else:
            return True
    if m<=0 :
        # arr[n][m] = False
        return False
    # if n < 1:
    #     return out1(x-10,n,m-1)
    # elif m <= 0:

        
    return out1(x//2+10,n-1,m) or out1(x-10,n,m-1)



def out():
    n, a,b,c,d = map(int,input().split())

    if c-d > n*(a+b) or n*(a-b) > c+d  :
        return "No"
    return "Yes"

    # for i in range(n//2+1):
    #     li.append(arr[i]+arr[n-i-1])
    
    # print(len(set(li))-1)



ans = 0 
for i in range(int(input())):
    x,y = map(int,input().split())

    a,b = map(int,input().split())
    print(min(b*min(x,y)+a*(max(x,y)-min(x,y)), a*(x+y) )) 
    