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 java.util.*;
import java.io.*;
public class rtz
{
    static long t;
    static long a,b,x,y;
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        t = sc.nextLong();
        for(int i=0;i<t;i++)
        {
            x = sc.nextLong();
            y = sc.nextLong();
            a = sc.nextLong();
            b = sc.nextLong();
            System.out.println(min(x,y));
        }
    }
    public static long min(long i,long j)
    {
        if(i==0)
        return j*a;
        else if(j==0)
        return i*a;
        else
        {
            if(i>0 && j>0)
            {
                if(b>2*a)
                return (i+j)*a;
                else
                {
                    if(i>=j)
                    return (b*j)+(a*(i-j));
                    else
                    return (b*i)+(a*(j-i));
                }
            }
            else if(i>0 && j<0)
            {
                return (a*(-1)*j)+(a*i);
            }
            else if(i<0 && j>0)
            {
                return (a*(-1)*i)+(a*j);
            }
            else
            {
                if(b>2*a)
                return (i+j)*a*(-1);
                else
                {
                    if(i>=j)
                    return (b*i*(-1))+(a*(i-j));
                    else
                    return (b*j*(-1))+(a*(j-i));
                }
            }
        }
    }
}