Screen resolution of Polycarp's monitor is a × b pixels. Unfortunately, there is one dead pixel at his screen. It has coordinates (x, y) (0 ≤ x < a, 0 ≤ y < b). You can consider columns of pixels to be numbered from 0 to a-1, and rows — from 0 to b-1.

Polycarp wants to open a rectangular window of maximal size, which doesn't contain the dead pixel. The boundaries of the window should be parallel to the sides of the screen.

Print the maximal area (in pixels) of a window that doesn't contain the dead pixel inside itself.

Input

In the first line you are given an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the test. In the next lines you are given descriptions of t test cases.

Each test case contains a single line which consists of 4 integers a, b, x and y (1 ≤ a, b ≤ 10^4; 0 ≤ x < a; 0 ≤ y < b) — the resolution of the screen and the coordinates of a dead pixel. It is guaranteed that a+b>2 (e.g. a=b=1 is impossible).

Output

Print t integers — the answers for each test case. Each answer should contain an integer equal to the maximal possible area (in pixels) of a rectangular window, that doesn't contain the dead pixel.

Example

Input


6
8 8 0 0
1 10 0 3
17 31 10 4
2 1 0 0
5 10 3 9
10 10 4 8


Output


56
6
442
1
45
80

Note

In the first test case, the screen resolution is 8 × 8, and the upper left pixel is a dead pixel. Here you can see one of two possible layouts of the maximal window.

<image>
import 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 heapPermutation(a, size, n): 
    global x
    # if size becomes 1 then prints the obtained 
    # permutation 
    if (size == 1):
        # printArr(a,n)
        x.append(a[:]) 
        return
  
    for i in range(size): 
        heapPermutation(a,size-1,n); 
  
        # if size is odd, swap first and last 
        # element 
        # else If size is even, swap ith and last element 
        if size&1: 
            a[0], a[size-1] = a[size-1],a[0] 
        else: 
            a[i], a[size-1] = a[size-1],a[i]
"""*******************************************************"""
x=[]
def main():
    t=inin()
    for _ in range(t):
        x,y,a,b=cin()
        p=max(a,x-a-1)
        q=max(b,y-b-1)
        # print(p,q)
        print(max(p*(y),q*(x)))
######## 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()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()