You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissions, out of which x have been successful. Thus, your current success rate on Codeforces is equal to x / y.

Your favorite rational number in the [0;1] range is p / q. Now you wonder: what is the smallest number of submissions you have to make if you want your success rate to be p / q?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each of the next t lines contains four integers x, y, p and q (0 ≤ x ≤ y ≤ 109; 0 ≤ p ≤ q ≤ 109; y > 0; q > 0).

It is guaranteed that p / q is an irreducible fraction.

Hacks. For hacks, an additional constraint of t ≤ 5 must be met.

Output

For each test case, output a single integer equal to the smallest number of submissions you have to make if you want your success rate to be equal to your favorite rational number, or -1 if this is impossible to achieve.

Example

Input

4
3 10 1 2
7 14 3 8
20 70 2 7
5 6 1 1


Output

4
10
0
-1

Note

In the first example, you have to make 4 successful submissions. Your success rate will be equal to 7 / 14, or 1 / 2.

In the second example, you have to make 2 successful and 8 unsuccessful submissions. Your success rate will be equal to 9 / 24, or 3 / 8.

In the third example, there is no need to make any new submissions. Your success rate is already equal to 20 / 70, or 2 / 7.

In the fourth example, the only unsuccessful submission breaks your hopes of having the success rate equal to 1.
import math
import sys
import bisect  # https://pythonworld.ru/moduli/modul-bisect.html
from heapq import heapify, heappop, heappush
from itertools import *  # https://pythonworld.ru/moduli/modul-itertools.html
from collections import deque, OrderedDict
sys.setrecursionlimit(10 ** 6)
# f = open('input.txt')
# f.close()
II = lambda: sys.stdin.readline()  # f.readline()
inp = lambda: int(II())
inpm = lambda: map(int, II().split())
inpl = lambda: list(inpm())
arr_mn = lambda _n, _m: [[0 for __ in range(_m)] for _ in range(_n)]
arr_nn = lambda _n: arr_mn(_n, _n)

EPS = 1e-9
INF = int(1e18)
MOD = int(1e9) + 7  # 998244353
N = 2000009
"""

"""
mod = 475
def check(p, s):
    i = (s // 50) % mod
    for _ in range(25):
        i = (i * 96 + 42) % mod
        if i + 26 == p:
            return True
    return False

def solve():
    x, y, p, q = inpm()
    if p == q and x != y:
        print(-1)
        return
    if p == 0:
        if x == 0:
            if y > 0:
                print(0)
            else:
                print(1)
        else:
            print(-1)
        return
    l = max(1, max((x + p - 1) // p , (y + q - 1) // q))
    r = 10 ** 9 + 1
    while r - l > 1:
        m = l + (r - l) // 2
        good = p * m - x
        all = q * m - y
        if all >= good:
            r = m
        else:
            l = m
    good = p * l - x
    all = q * l - y
    if all >= good:
        print(all)
    else:
        print(q * r - y)
'''
1
0 0 4 7

3
3 10 1 2
7 14 3 8
20 70 2 7

4
3 10 1 2
7 14 3 8
20 70 2 7
5 6 1 1

'''

def main():
    t = inp()  # 1  #
    for i in range(t):
        solve()
        # print()
main()
