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):
        x = 0
        y = 1
        return b, x, y

    d, x1, y1 = gcd(b % a, a)
    x = y1 - (b // a) * x1
    y = x1
    return d, x, y

t = int(input())

for _ in range(t):
    inp = input().split()
    x, y, p, q = int(inp[0]), int(inp[1]), int(inp[2]), int(inp[3])

    if(p == q):
        if(x == y):
            print(0)
        else:
            print(-1)
        continue

    if(p == 0):
        if(x == 0):
            print(0)
        else:
            print(-1)
        continue

    c = (y*p) - (x*q)
    # print("c = {}".format(c))

    p = -p;
    g, d1, d2 = gcd(q, p);

    if(g < 0):
        c *= -1;
        g *= -1;

    d1 *= c;
    d2 *= c;

    # print("q = {} d1 = {} + p = {} d2 = {}".format(q, d1, p, d2))
    if(d1 < 0):
        val = abs(d1);
        t = abs(val//abs(p));

        if( abs(t*(p)) < val):
          t+=1;
        # print("t<0 = {}".format(t))

        d1 -= t*(p);
        d2 += t*(q);

    elif(d1 > 0):
        val = abs(d1);
        t = abs(val//abs(p));

        # print("t>0 = {}".format(t))

        d1 += t*(p);
        d2 -= t*(q);

    if(d1 > d2):
        t = (d1 - d2)//(p + q);

        d1 -= t*(p);
        d2 += t*(q);

    if(d1 > d2):
        d1 -= (p);
        d2 += (q);

    print(d2)

    # print("q = {} d1 = {} + p = {} d2 = {}".format(q, d1, p, d2))
