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 math import radians
from heapq import heapify, heappush, heappop
import bisect
from math import pi
from collections import deque
from math import factorial
from math import log, ceil
from collections import defaultdict
from math import *
from sys import stdin, stdout
import itertools
import os
import sys
import threading
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import *
# from math import ceil, floor, log, sqrt, factorial, pow, pi, gcd
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
from fractions import Fraction
mod = int(pow(10, 9)+7)
# mod = 998244353


def ii(): return int(input())


def si(): return str(input())


def mi(): return map(int, input().split())


def li1(): return list(mi())


def fii(): return int(stdin.readline())


def fsi(): return str(stdin.readline())


def fmi(): return map(int, stdin.readline().split())


def fli(): return list(fmi())


abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
       'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
       'z': 25}


def getKey(item): return item[0]


def sort2(l): return sorted(l, key=getKey)


def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]


def isPowerOfTwo(x): return (x and (not (x & (x - 1))))


def decimalToBinary(n): return bin(n).replace("0b", "")


def ntl(n): return [int(i) for i in str(n)]


def powerMod(x, y, p):
    res = 1
    x %= p
    while y > 0:
        if y & 1:
            res = (res * x) % p
        y = y >> 1
        x = (x * x) % p
    return res


graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000


def bfs(d, v):
    q = []
    q.append(v)
    visited[v] = 1
    while len(q) != 0:
        x = q[0]
        q.pop(0)
        for i in d[x]:
            if visited[i] != 1:
                visited[i] = 1
                q.append(i)
        print(x)


def make_graph(e):
    d = {}
    for i in range(e):
        x, y = mi()
        if x not in d:
            d[x] = [y]
        else:
            d[x].append(y)
        if y not in d:
            d[y] = [x]
        else:
            d[y].append(x)
    return d


def gr2(n):
    d = defaultdict(list)
    for i in range(n):
        x, y = mi()
        d[x].append(y)
    return d


def connected_components(graph):
    seen = set()

    def dfs(v):
        vs = set([v])
        component = []
        while vs:
            v = vs.pop()
            seen.add(v)
            vs |= set(graph[v]) - seen
            component.append(v)
        return component

    ans = []
    for v in graph:
        if v not in seen:
            d = dfs(v)
            ans.append(d)
    return ans


def primeFactors(n):
    s = set()
    while n % 2 == 0:
        s.add(2)
        n = n // 2
    for i in range(3, int(sqrt(n)) + 1, 2):
        while n % i == 0:
            s.add(i)
            n = n // i
    if n > 2:
        s.add(n)
    return s


def find_all(a_str, sub):
    start = 0
    while True:
        start = a_str.find(sub, start)
        if start == -1:
            return
        yield start
        start += len(sub)


def SieveOfEratosthenes(n, isPrime):
    isPrime[0] = isPrime[1] = False
    for i in range(2, n):
        isPrime[i] = True
    p = 2
    while (p * p <= n):
        if (isPrime[p] == True):
            i = p * p
            while (i <= n):
                isPrime[i] = False
                i += p
        p += 1
    return isPrime


def dijkstra(edges, f, t):
    g = defaultdict(list)
    for l, r, c in edges:
        g[l].append((c, r))

    q, seen, mins = [(0, f, ())], set(), {f: 0}
    while q:
        (cost, v1, path) = heappop(q)
        if v1 not in seen:
            seen.add(v1)
            path = (v1, path)
            if v1 == t:
                return (cost, path)

            for c, v2 in g.get(v1, ()):
                if v2 in seen:
                    continue
                prev = mins.get(v2, None)
                next = cost + c
                if prev is None or next < prev:
                    mins[v2] = next
                    heappush(q, (next, v2, path))
    return float("inf")


def binsearch(a, l, r, x):
    while l <= r:
        mid = l + (r-1)//2
        if a[mid]:
            return mid
        elif a[mid] > x:
            l = mid-1
        else:
            r = mid+1
    return -1


# def input():
#    return stdin.buffer.readline()


def readTree(n):
    adj = [set() for _ in range(n)]
    for _ in range(n-1):
        u, v = map(int, input().split())
        adj[u-1].add(v-1)
        adj[v-1].add(u-1)
    return adj


def treeOrderByDepth(n, adj, root=0):
    parent = [-2] + [-1]*(n-1)
    ordered = []
    q = deque()
    q.append(root)
    depth = [0] * n
    while q:
        c = q.popleft()
        ordered.append(c)
        for a in adj[c]:
            if parent[a] == -1:
                parent[a] = c
                depth[a] = depth[c] + 1
                q.append(a)
    return (ordered, parent, depth)

for _ in range(ii()):
    a, b = mi()
    x = min(a, b)
    print(max(a,b,2*x)**2)