# n(A+X)=B+X
# B-A = (n-1)(A+X)
# n-1=(B-A)/(A+X)<=(B-A)/Aより、B-Aの約数であって(B-A)/A以下な最大のn-1を探せばいい？

# 約数列挙O(√n)
def make_divisors(n):
    lower_divisors , upper_divisors = [], []
    i = 1
    while i*i <= n:
        if n % i == 0:
            lower_divisors.append(i)
            if i != n // i:
                upper_divisors.append(n//i)
        i += 1
    return lower_divisors + upper_divisors[::-1]

T = int(input())
for _ in range(T):
    A, B = map(int, input().split())
    div = make_divisors(B-A)
    ans = -1
    for d in div[::-1]:
        if (B-A)/A<d:
            continue
        n = d+1
        ans = (B-n*A)/(n-1)
        break
    print(int(ans))
"🐬"=="🐬"