Find the minimum area of a square land on which you can place two identical rectangular a × b houses. The sides of the houses should be parallel to the sides of the desired square land.

Formally, 

  * You are given two identical rectangles with side lengths a and b (1 ≤ a, b ≤ 100) — positive integers (you are given just the sizes, but not their positions). 
  * Find the square of the minimum area that contains both given rectangles. Rectangles can be rotated (both or just one), moved, but the sides of the rectangles should be parallel to the sides of the desired square. 



Two rectangles can touch each other (side or corner), but cannot intersect. Rectangles can also touch the sides of the square but must be completely inside it. You can rotate the rectangles. Take a look at the examples for a better understanding.

<image> The picture shows a square that contains red and green rectangles.

Input

The first line contains an integer t (1 ≤ t ≤ 10 000) —the number of test cases in the input. Then t test cases follow.

Each test case is a line containing two integers a, b (1 ≤ a, b ≤ 100) — side lengths of the rectangles.

Output

Print t answers to the test cases. Each answer must be a single integer — minimal area of square land, that contains two rectangles with dimensions a × b.

Example

Input


8
3 2
4 2
1 1
3 1
4 7
1 3
7 4
100 100


Output


16
16
4
9
64
9
64
40000

Note

Below are the answers for the first two test cases: 

<image> <image>
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**7)
# 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
q=[]       
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j
from collections import defaultdict 
  
# Function to find number of subarrays   
# with sum exactly equal to k.  
def findSubarraySum(arr, n, Sum):  
   
    # Dictionary to store number of subarrays  
    # starting from index zero having   
    # particular value of sum.  
    prevSum = defaultdict(lambda : 0) 
    
    res = 0 
    
    # Sum of elements so far.  
    currsum = 0 
    
    for i in range(0, n):   
    
        # Add current element to sum so far.  
        currsum += arr[i]  
    
        # If currsum is equal to desired sum,  
        # then a new subarray is found. So  
        # increase count of subarrays.  
        if currsum == Sum:   
            res += 1         
    
        # currsum exceeds given sum by currsum  - sum. 
        # Find number of subarrays having   
        # this sum and exclude those subarrays  
        # from currsum by increasing count by   
        # same amount.  
        if (currsum - Sum) in prevSum: 
            res += prevSum[currsum - Sum]  
            
    
        # Add currsum value to count of   
        # different values of sum.  
        prevSum[currsum] += 1 
       
    return res 
import random
"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        n,m=cin()
        x=min(n,m)
        y=max(n,m)
        x*=2
        x=max(x,y)
        print(x*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'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()