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.
INF = 10**40

def gcd_ex(A, B):
    if A == 0:
        return B, 0, 1
    g, a1, b1 = gcd_ex(B % A, A)
    a = b1 - (B // A) * a1
    b = a1
    return g, a, b

def round_down(a, b):
    s = (a < 0) != (b < 0)
    a = abs(a)
    b = abs(b)
    if not s:
        return a // b
    return -((a + b - 1) // b)

def round_up(a, b):
    s = (a < 0) != (b < 0)
    a = abs(a)
    b = abs(b)
    if not s:
        return (a + b - 1) // b
    return -(a // b)

def solve():
    x, y, p, q = map(int, input().split())

    A = q
    B = -p
    C = p * y - q * x

    g, a0, b0 = gcd_ex(A, -B)
    b0 *= -1

    if C % g != 0:
        print(-1)
        return
    a0 *= C // g
    b0 *= C // g

    k_up1 = 0
    if B == 0:
        if -a0 > 0:
            print(-1)
            return
        k_up1 = INF
    else:
        k_up1 = round_down(-a0, B // g)

    k_down2 = 0
    k_up2 = 0
    if A + B == 0:
        if b0 - a0 < 0:
            print(-1)
            return
        k_down2 = -INF
        k_up2 = INF
    elif A + B > 0:
        k_down2 = -INF
        k_up2 = round_down(b0 - a0, (A + B) // g)
    else:
        k_down2 = round_up(b0 - a0, (A + B) // g)
        k_up2 = INF

    k_down = k_down2
    k_up = min(k_up1, k_up2)
    if k_down > k_up:
        print(-1)
        return

    if k_up == INF:
        raise Exception
    b = b0 - (A // g) * k_up

    print(b)

def main():
    t = int(input())
    for i in range(t):
        solve()

main()
