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.
#                               #
    # author : samars_diary #
    # 17:02:07 | 25-09-2020 #
#                               #

import sys, os.path

if(os.path.exists('input.txt')):
    sys.stdin = open('input.txt',"r")
    sys.stdout = open('output.txt',"w")

sys.setrecursionlimit(10 ** 5)

def i(): return sys.stdin.readline().strip()
def ii(): return int(sys.stdin.readline())
def li(): return list(sys.stdin.readline().strip())
def mii(): return map(int, sys.stdin.readline().split())
def lii(): return list(map(int, sys.stdin.readline().strip().split()))

#print=sys.stdout.write

def solve():

    a, b = mii() ; c, d = mii() ; x, y = min(abs(a), abs(b)), max(abs(a), abs(b))
    if (a>=0 and b<=0) or (a<=0 and b>=0): print(min((x+y)*c, (max(x, y)*d)+(x+y)*c))
    else: print(min((x+y)*c, d*y+abs(x-y)*c, d*x+(y-x)*c))

for _ in range(ii()): solve()