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.
#qn given an arr of size n for every  i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
import sys
input = sys.stdin.readline
 #qn given an arr of size n for every  i position find a min j such that j>i and arr[j]>arr[i]
#it prints the position of the nearest .
# import sys
import heapq 
import copy
import math
#heapq.heapify(li) 
#
#heapq.heappush(li,4) 
#
#heapq.heappop(li)
#
# &	Bitwise AND Operator	10 & 7 = 2
# |	Bitwise OR Operator	10 | 7 = 15
# ^	Bitwise XOR Operator	10 ^ 7 = 13

# <<	Bitwise Left Shift operator	10<<2 = 40
# >>	Bitwise Right Shift Operator
############ ---- Input Functions ---- #######Start#####
def inp():
	return(int(input()))
def inlt(): 
	return(list(map(int,input().split())))
def insr():
	s = input()
	return(list(s[:len(s) - 1]))
def invr():
	return(map(int,input().split()))
 ############ ---- Input Functions ---- #######End#####   
def bin_search(arr,l,r,val):#strickly greater
	if arr[r] <= val:
		return r+1	
	if r-l < 2:
		if arr[l]>val:
			return l
		else:
			return r
	mid = int((l+r)/2)
	if arr[mid] <= val:
		return bin_search(arr,mid,r,val)
	else:
		return bin_search(arr,l,mid,val)

def pr_list(a):
	print(*a, sep=" ")
def main():
	tests =  inp()
	mod = 1000000007
	limit = 10**18
	# print(limit)
	for test in range(tests):
		[x,y] =  inlt()
		[a,b] = inlt()
		if b>2*a:
			print(a*(x + y))
		else:
			X = min(x,y)
			Y = max(x,y)
			print(b*(X) + a*(Y-X))
			



	return 0


# if for two position i,j such that j<i and arr[j]>arr[i] I wont need to store the position i as I wont requre it for any further computation. I cound simply use i instead of j.
#to do that i main tain a stack which is initiallized with the last elment and I traverse the arr backwords because for any ith entry the answer does not depends on 0 to i-1.
# for every element I would continue to pop the stack (from last) till i recieve an elemnt larger than current one. The elemnt which i poped are not needed as i could use current elemnt.
# the order is n as an element only enters the stack ones.
if __name__== "__main__":
  main()