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.
from sys import stdin

def gcd(a,b, xy):
    if (a==0):
        xy[0]=0
        xy[1]=1
        return b
    newxy = [0,0]
    d = gcd(b%a,a,newxy)
    xy[0]=newxy[1]-(b/a)*newxy[0]
    xy[1]=newxy[0]
    return d

def solve_diofant(a,b,c, solutionList):
    g = gcd(abs(a), abs(b), solutionList)
    if c%g != 0: 
        return g
    solutionList[0] *= c/g;
    solutionList[1] *= c/g;
    if a < 0:
        solutionList[0] *= -1
    if b < 0:
        solutionList[1] *= -1
    return g

def divRoundUp1(p, q):
    print str("DivRoundUp1 ") + str(p) + " " + str(q)
    if q < 0:
        p = -p
        q = -q
    if p < 0:
        #print str(p/q)
        return p / q
    else:
        #print str( (p+q-1)/q)
        return (p + q - 1) / q

def divRoundUp(p, q):
    #print str("DivRoundUp ") + str((p+q-1)/q)
    return (p+q-1)/q


def solve(x, y, p, q):
    if p == 0:
        if x == 0:
            return 0
        else:
            return -1
    elif p == q:
        if x == y:
            return 0
        else:
            return -1

    diofSol = [0,0]
    g = solve_diofant(q, -p, y*p - x*q, diofSol)
    
    a = diofSol[0]
    b = diofSol[1]
    
    #print str("diofant sol: ") + str(a) + " " + str(b)
    
    k = max(divRoundUp(-a, p), divRoundUp((a - b), (q - p)))
    return b + k*q

[n] = map(int, stdin.readline().split())

for i in xrange(n):
    x,y,p,q = map(int, stdin.readline().split())
    print solve(x,y,p,q)
