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 Ans(k):

    x, y, p, q = k[0], k[1], k[2], k[3]
    if (p/q == 1 and x != y) or (p == 0 and x!= 0):
        print(-1)
    elif p == 0 and x == 0:
        print(0)
    else:
        up = 10**9
        down = 1
        dy = y-x
        while (up > down + 1):
            mid = (up+down)//2
##            print(" up: " + str(up) + " mid: " + str(mid) + " down: " + str(down) )
            qf = q*mid
            pf = p*mid
            
            dq = qf-pf
        
            if( qf-y >= dq-dy and qf >= y and dq >= dy):
                up = mid
            else:
                down = mid
        if( q*down-y >= (q-p)*down-dy and q*down >= y and (q-p)*down >= dy):
            print(q*down - y)
        elif q*mid-y >= (q-p)*mid-dy and q*mid >= y and (q-p)*mid>=dy:
            print(q*mid - y)
        else:
            print(q*up - y)
            
            

t = int( input() )
k = [0]*t
for i in range(t):
    k[i] = list(map(int, input().split() ))
for i in range(t):
    Ans(k[i])

