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.
def gcd(a, b):
    if a == 0:
        return b
    return gcd(b % a, a)


def get(x, y, p, q):
    low, high, mid, best = 1, 1 << 65, 0, -1
    while low <= high:
        mid = (low + high + 1) >> 1
        tp = p * mid
        tq = q * mid
        if tq >= y and x <= tp <= x + tq - y:
            best = mid
            high = mid - 1
        else:
            low = mid + 1
    return best

T = int(input())
for _ in range(T):
    x, y, p, q = map(int, input().split())
    if x == 0 and p == q:
        print(-1)
        continue
    if x == y and p == 0:
        print(-1)
        continue
    if x == 0 and p == 0:
        print(0)
        continue
    nx, ny = x, y
    nx //= gcd(x, y)
    ny //= gcd(x, y)
    #print("reduced", nx, ny)
    need = get(nx, ny, p, q)
    rx = need * p
    ry = need * q
    #print("first new", rx, ry)
    low, high, mid, best = 0, 1 << 63, 0, -1
    while low <= high:
        mid = (low + high + 1) >> 1
        if rx + mid * p >= x and ry + mid * q >= y and ry + mid * q - y >= rx + mid * p - x:
            best = mid
            high = mid - 1
        else:
            low = mid + 1
    if best == -1:
        print(-1)
        continue
    rx += best * p
    ry += best * q
    #print("resulting", rx, ry)
    if rx * q == ry * p:
        print(ry - y)
    else:
        print(-1)

