做法是对 $a,b$ 分解因数，然后计算出 $ab$ 的因数。
然后遍历 $ab$ 的因数作为 $x$，$O(1)$ 计算出对应的 $y$。

复杂度是 $ab$ 的因数数量，1e6 级别，再乘上 T 就是 1e7。按理说能过。

然后 TLE on test25 了。

import sys
import math
input = sys.stdin.readline
def great_sqrt(x):
    _ = int(math.sqrt(x))
    while _*_ < x:
        _+=1
    while _*_ > x:
        _-=1
    return _
if __name__ == "__main__":
    t = int(input())
    while t!=0:
        a,b,c,d=map(int,input().split())
        _divs=[]
        _divs2=[]
        for i in range(1,great_sqrt(a)+1):
            if a%i==0:
                _divs.append(i)
                if a/i != i:
                    _divs.append(a/i)
        for i in range(1,great_sqrt(b)+1):
            if b%i==0:
                _divs2.append(i)
                if b/i != i:
                    _divs2.append(b/i)
        divs = []
        for div in _divs:
            for div2 in _divs2:
                divs.append(div*div2)
      #          print("(%d,%d)"%(div,div2))
        divs.append(c)
        print(len(divs))
        ok=0
        for x in divs:
            x_tmp = x
            if (a-1)//x_tmp != c//x_tmp:
                x = ((a-1)//x_tmp)*x_tmp + x_tmp
            #    print("tempy = %d"%(y))
                if x+x_tmp <= c:
                    x=x+x_tmp
                if x <= a or x > c:
                    continue
            if x<=a or x > c:
                continue
       #     print("x = %d"%(x))
           # print("y_tmp = %d"%(y_tmp))
            if x==c:
                y_tmp=a*b
            else:
                y_tmp = a*b/int(x)
            if (b-1)//y_tmp != d//y_tmp:
                y = ((b-1)//y_tmp)*y_tmp + y_tmp
            #    print("tempy = %d"%(y))
                if y+y_tmp <= d:
                    y=y+y_tmp
                if y <= b or y > d:
                    continue
                print("%d %d"%(x,y),end='\n')
                ok=1
                break
        if ok==0:
            print("-1 -1",end='\n')
      #  print("=====================")
        t-=1