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 sys
import math


def gcd(a, b, x, y):
    if a == 0:
        x[0], y[0] = 0, 1
        return b
    x1, y1 = [0], [0]
    d = gcd(b % a, a, x1, y1)
    x[0] = y1[0] - (b / a) * x1[0]
    y[0] = x1[0]
    return d


def check(a, b, p, q, k):
    return a + p * k >= 0 and b + q * k >= 0


def ceil(a, b):
    assert b > 0
    if a > 0:
        return (a + b - 1) / b
    else:
        return -(-a / b)


def get_answer(a, b, p, q):
    k1 = ceil(a - b, q - p)
    k2 = ceil(-a, p)
    # print("k1 = {}, k2 = {}\n".format(k1, k2))
    return b + q * max(k1, k2)


def solve(x, y, p, q):
    if p == q:
        return 0 if x == y else -1
    if p == 0:
        return 0 if x == 0 else -1
    c = y * p - x * q
    a, b = [0], [0]
    gcd(q, p, a, b)
    a, b = a[0], -b[0]
    a *= c
    b *= c
    # print("a = {}, b = {}, c = {}, p = {}, q = {}".format(a, b, c, p, q))
    return get_answer(a, b, p, q)


fin = sys.stdin
fout = sys.stdout

# fin = open("C.in", "r")
# fout = open("C.out", "w")


t = int(fin.readline())
for _ in range(t):
    x, y, p, q = map(int, fin.readline().split())
    fout.write("{}\n".format(solve(x, y, p, q)))
