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;
const long long mod = 1e9 + 7;
long long expo(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x % mod) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
long long inv(long long x) { return expo(x, mod - 2); }
long long add(long long x, long long y) { return (x + y) % mod; }
long long sub(long long x, long long y) { return (x - y) % mod; }
long long mul(long long x, long long y) {
  return ((x % mod) * (y % mod)) % mod;
}
long long power(long long x, long long y) {
  long long val = 2;
  for (long long i = 1; i < y; i++) val = mul(val, 2);
  return val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long x, y, a, b;
    cin >> x >> y >> a >> b;
    long long a1 = (x + y) * a;
    long long a2 = min(x, y) * b + abs(x - y) * a;
    cout << min(a1, a2);
    cout << "\n";
  }
}
