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.
n=int(input())
a=[]
for i in range (n*2):
    t=input().split()
    a.append(t)
for l in range (n):  
    money1=0
    money2=0
    
    if (2*int(a[(l*2)+1][0]))>=(int(a[(l*2)+1][1])):
        if int(a[l*2][1]) != int(a[l*2][0]):
            
            m=int(a[l*2][1])-int(a[l*2][0])
            if (m > 0):
                t=int(a[l*2][0])
            else:
                t=int(a[l*2][1])
            money1=abs(m)*int(a[(l*2)+1][0])
            if (int(a[l*2][1]==0)):
                money2=0
            else:
                mn=int(a[l*2][0])-int(a[l*2][1])
                money2=int(a[(l*2)+1][1])*t
            sum1=money1+money2
            print(sum1)
           
        else: 
            (int(a[l*2][1])==int(a[l*2][0]))
            if (int(a[l*2][1]==0)):
                money2=0
            else:
                if (int(a[l*2][1])!=0):
                    money2=int(a[(l*2)+1][1]) * int(a[l*2][0]) 
            sum1=money1+money2
            print(sum1)
           
            
    else :
        if (2*(int(a[(l*2)+1][0])) <=(int(a[(l*2)+1][1]))):
            if int(a[l*2][1]) != int(a[l*2][0]):
                mmm= int(a[l*2][1])-int(a[l*2][0])
                if (mmm > 0):
                    t=int(a[l*2][0])
                else:
                    t=int(a[l*2][1])
                money1=abs(mmm)*int(a[(l*2)+1][0])
                if (int(a[l*2][0]==0)):
                    money2=0
                else:
                    money2=2* t * int(a[(l*2)+1][0])
                sum1=money1+money2
                print(sum1)
            else:
                if (int(a[l*2][1])==int(a[l*2][0])):
                    if (int(a[l*2][0]==0)):
                        money2=0
                    else:
                        (int(a[l*2][0])!=0)
                        
                        money2=2*int(a[(l*2)+1][0])*int(a[l*2][0])
                    sum1=money1+money2
                    print(sum1)
