Wabbit is trying to move a box containing food for the rest of the zoo in the coordinate plane from the point (x_1,y_1) to the point (x_2,y_2).

He has a rope, which he can use to pull the box. He can only pull the box if he stands exactly 1 unit away from the box in the direction of one of two coordinate axes. He will pull the box to where he is standing before moving out of the way in the same direction by 1 unit. 

<image>

For example, if the box is at the point (1,2) and Wabbit is standing at the point (2,2), he can pull the box right by 1 unit, with the box ending up at the point (2,2) and Wabbit ending at the point (3,2).

Also, Wabbit can move 1 unit to the right, left, up, or down without pulling the box. In this case, it is not necessary for him to be in exactly 1 unit away from the box. If he wants to pull the box again, he must return to a point next to the box. Also, Wabbit can't move to the point where the box is located.

Wabbit can start at any point. It takes 1 second to travel 1 unit right, left, up, or down, regardless of whether he pulls the box while moving.

Determine the minimum amount of time he needs to move the box from (x_1,y_1) to (x_2,y_2). Note that the point where Wabbit ends up at does not matter.

Input

Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000): the number of test cases. The description of the test cases follows.

Each of the next t lines contains four space-separated integers x_1, y_1, x_2, y_2 (1 ≤ x_1, y_1, x_2, y_2 ≤ 10^9), describing the next test case.

Output

For each test case, print a single integer: the minimum time in seconds Wabbit needs to bring the box from (x_1,y_1) to (x_2,y_2).

Example

Input


2
1 2 2 2
1 1 2 2


Output


1
4

Note

In the first test case, the starting and the ending points of the box are (1,2) and (2,2) respectively. This is the same as the picture in the statement. Wabbit needs only 1 second to move as shown in the picture in the statement.

In the second test case, Wabbit can start at the point (2,1). He pulls the box to (2,1) while moving to (3,1). He then moves to (3,2) and then to (2,2) without pulling the box. Then, he pulls the box to (2,2) while moving to (2,3). It takes 4 seconds.
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
import sys
sys.setrecursionlimit(10**4)
# sys.stdin = open('inpy.txt', 'r')
# sys.stdout = open('outpy.txt', 'w')
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
i_m=9223372036854775807
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 GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
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
prime=[]
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    # Print the number of two's that divide n 
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2
          
    # n must be odd at this point 
    # so a skip of 2 ( i = i + 2) can be used 
    for i in range(3,int(math.sqrt(n))+1,2): 
          
        # while i divides n , print i ad divide n 
        while n % i== 0: 
            a.append(i) 
            n = n // i 
              
    # Condition if n is a prime 
    # number greater than 2 
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        a,b,c,d=cin()
        a=abs(c-a)
        b=abs(b-d)
        ans=a+b
        if a!=0 and b!=0:
            ans+=2
        print(ans)
        


        
    
######## 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 yferent 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'R' [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'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()