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.
# -*- coding: utf-8 -*-
"""
Created on Mon Apr 27 16:49:13 2020

@author: alexi
"""



#http://codeforces.com/contest/1342/problem/A --- Alexis Galvan 


def road_zero():
    
    cases = int(input())
    numbers = []
    prices = []
    
    for i in range(cases):
        numbers.append(list(map(int, input().split())))
        prices.append(list(map(int, input().split())))
    
    for i in range(len(numbers)):
        if prices[i][0] == 0:
            print(0)
            continue
        
        output = 0
        minimum = min(abs(numbers[i][0]),abs(numbers[i][1]))
        best = min((minimum*prices[i][0])*2,minimum*prices[i][1])
        
        if abs(numbers[i][0]) == minimum:
            numbers[i][0] = 0
        else:
            numbers[i][0] -= minimum
            
        if abs(numbers[i][1]) == minimum:
            numbers[i][1] = 0
        else:
            numbers[i][1] -= minimum
        
        output += best
        
        if numbers[i][0] == 0 and numbers[i][1] == 0:
            print(output)
            continue
        
        maximum = max(numbers[i][0],numbers[i][1])
        output += (maximum*(prices[i][0]))
        
        print(output)
        
road_zero()
        