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 sys


def find_cost(case):
    x, y, a, b = case["x"], case["y"], case["a"], case["b"]
    res = 0
    if x == 0 and y == 0:
        return res
    if x == 0:
        res = count_single(y, a)
    elif y == 0:
        res = count_single(x, a)
    elif 2 * a <= b or x * y < 0:
        res = count_single(x, a) + count_single(y, a)
    elif x == y:
        res = count_double(x, b)
    else:
        max_val, min_val = max(x, y), min(x, y)
        diff = max_val - min_val
        res = diff * a + (max_val - diff) * b

    return res


def count_single(x, a):
    res = abs(x) * a
    return res

def count_double(x, b):
    res = abs(x) * b
    return res
    
inp = sys.stdin.read()
inp = inp.split("\n")[1:]
cases = []
for n in range(len(inp))[:-1:2]:
    x, y = inp[n].split()
    a, b = inp[n+1].split()
    cases.append({"x": int(x), "y": int(y), "a": int(a), "b": int(b)})

for case in cases:
    res = find_cost(case)
    print(res)

