human
Being tired of participating in too many Codeforces rounds, Gildong decided to take some rest in a park. He sat down on a bench, and soon he found two rabbits hopping around. One of the rabbits was taller than the other.

He noticed that the two rabbits were hopping towards each other. The positions of the two rabbits can be represented as integer coordinates on a horizontal line. The taller rabbit is currently on position x, and the shorter rabbit is currently on position y (x < y). Every second, each rabbit hops to another position. The taller rabbit hops to the positive direction by a, and the shorter rabbit hops to the negative direction by b.


For example, let's say x=0, y=10, a=2, and b=3. At the 1-st second, each rabbit will be at position 2 and 7. At the 2-nd second, both rabbits will be at position 4.

Gildong is now wondering: Will the two rabbits be at the same position at the same moment? If so, how long will it take? Let's find a moment in time (in seconds) after which the rabbits will be at the same point.

Input

Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000).

Each test case contains exactly one line. The line consists of four integers x, y, a, b (0 ≤ x < y ≤ 10^9, 1 ≤ a,b ≤ 10^9) — the current position of the taller rabbit, the current position of the shorter rabbit, the hopping distance of the taller rabbit, and the hopping distance of the shorter rabbit, respectively.

Output

For each test case, print the single integer: number of seconds the two rabbits will take to be at the same position.

If the two rabbits will never be at the same position simultaneously, print -1.

Example

Input

5
0 10 2 3
0 10 3 3
900000000 1000000000 1 9999999
1 2 1 1
1 3 1 1

Output

2
-1
10
-1
1

Note

The first case is explained in the description.

In the second case, each rabbit will be at position 3 and 7 respectively at the 1-st second. But in the 2-nd second they will be at 6 and 4 respectively, and we can see that they will never...
impor ... Write me a Python program to find a moment in time (in seconds) after which the rabbits will be at the same point?
assistant
#Here is Python program to find a moment in time (in seconds) after which the rabbits will be at the same point.
t sys 
# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l

"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        a,b,x,y=cin()
        x=y+x
        b=b-a
        if(b%x==0):
            print(b//x)
        else:
            print(-1)
        
        
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
from io import IOBase, BytesIO
 
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
  ...