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.

//1342A
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Road_To_Zero {
    public static void main(String[] args) throws IOException {
        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
        String test_numS=bf.readLine();
        int test_num=Integer.parseInt(test_numS);
        while(test_num>0){
            String XandY=bf.readLine();
            String []xy=XandY.split(" ");
            int x=Integer.parseInt(xy[0]);
            int y=Integer.parseInt(xy[1]);
            String AandB=bf.readLine();
            String ab[]=AandB.split(" ");
            int a=Integer.parseInt(ab[0]);
            int b=Integer.parseInt(ab[1]);
            solve(x,y,a,b);

            test_num--;
        }
        bf.close();
    }
    public static void solve(int x,int y,int a,int b){
        long sum=0;
        long lx=x,ly=y,la=a,lb=b;
        if (2*la<=lb){//当a的2倍仍小于b时，可全用a一个一个消除
            sum=(lx+ly)*la;
            System.out.println(sum);
        }else if (2*la>lb){
            if (lx>ly){
                sum+=(lx-ly)*la;
                sum+=ly*lb;
                System.out.println(sum);
            }else if (lx<ly){
                sum+=(ly-lx)*la;
                sum+=lx*lb;
                System.out.println(sum);
            }else if (lx==ly){
                sum+=lx*lb;
                System.out.println(sum);
            }
        }
    }
}
