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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  long long arr[4 * n], brr[n], b, c, d, e, f, g, j = 0;
  for (int i = 0; i < (4 * n); i += 4) {
    cin >> arr[i] >> arr[i + 1] >> arr[i + 2] >> arr[i + 3];
    c = arr[i];
    d = arr[i + 1];
    b = arr[i + 2] * 2;
    e = arr[i + 2];
    f = arr[i + 3];
    if (c == d) {
      if (b >= f) {
        g = f * c;
      } else {
        g = b * c;
      }
    } else if (c > d) {
      if (b >= f) {
        g = (f * d) + (c - d) * e;
      } else {
        g = (c + d) * e;
      }
    } else if (c < d) {
      if (b >= f) {
        g = (f * c) + (d - c) * e;
      } else {
        g = (c + d) * e;
      }
    }
    brr[j] = g;
    j++;
  }
  for (int i = 0; i < n; i++) {
    printf("%lld\n", brr[i]);
  }
}
