Gottfried learned about binary number representation. He then came up with this task and presented it to you.

You are given a collection of n non-negative integers a_1, …, a_n. You are allowed to perform the following operation: choose two distinct indices 1 ≤ i, j ≤ n. If before the operation a_i = x, a_j = y, then after the operation a_i = x~AND~y, a_j = x~OR~y, where AND and OR are bitwise AND and OR respectively (refer to the Notes section for formal description). The operation may be performed any number of times (possibly zero).

After all operations are done, compute ∑_{i=1}^n a_i^2 — the sum of squares of all a_i. What is the largest sum of squares you can achieve?

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).

The second line contains n integers a_1, …, a_n (0 ≤ a_i < 2^{20}).

Output

Print a single integer — the largest possible sum of squares that can be achieved after several (possibly zero) operations.

Examples

Input


1
123


Output


15129


Input


3
1 3 5


Output


51


Input


2
349525 699050


Output


1099509530625

Note

In the first sample no operation can be made, thus the answer is 123^2.

In the second sample we can obtain the collection 1, 1, 7, and 1^2 + 1^2 + 7^2 = 51.

If x and y are represented in binary with equal number of bits (possibly with leading zeros), then each bit of x~AND~y is set to 1 if and only if both corresponding bits of x and y are set to 1. Similarly, each bit of x~OR~y is set to 1 if and only if at least one of the corresponding bits of x and y are set to 1. For example, x = 3 and y = 5 are represented as 011_2 and 101_2 (highest bit first). Then, x~AND~y = 001_2 = 1, and x~OR~y = 111_2 = 7.


# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
	f = open('q1.input', 'r')
except IOError:
	f = sys.stdin
if 'xrange' in dir(__builtins__):
	range = xrange
# print(f.readline())
sys.setrecursionlimit(10**2)

	
def print_case_iterable(case_num, iterable):
	print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))

def print_case_number(case_num, iterable):
	print("Case #{}: {}".format(case_num,iterable))

def print_iterable(A):
	print (' '.join(A))

def read_int():
	return int(f.readline().strip())
def read_int_array():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def read_string():
	return list(f.readline().strip())
def ri():
	return int(f.readline().strip())
def ria():
	return [int(x) for x in f.readline().strip().split(" ")]
def rns():
	a =  [x for x in f.readline().split(" ")]
	return int(a[0]), a[1].strip()
def rs():
	return list(f.readline().strip())
def bi(x):
	return bin(x)[2:]





from collections import deque
import math
# NUMBER = 10**9 + 7
NUMBER = 998244353
def factorial(n) : 
	M = NUMBER
	f = 1
  
	for i in range(1, n + 1):  
		f = (f * i) % M # Now f never can  
						# exceed 10^9+7  
	return f
def mult(a,b):
	return (a * b) % NUMBER

def minus(a , b):
	return (a - b) % NUMBER

def plus(a , b):
	return (a + b) % NUMBER

def egcd(a, b):
	if a == 0:
		return (b, 0, 1)
	else:
		g, y, x = egcd(b % a, a)
		return (g, x - (b // a) * y, y)

def modinv(a):
	m = NUMBER
	g, x, y = egcd(a, m)
	if g != 1:
		raise Exception('modular inverse does not exist')
	else:
		return x % m
def choose(n,k):
	if n < k:
		assert false
	return  mult(factorial(n), modinv(mult(factorial(k),factorial(n-k))))
from collections import deque, defaultdict 
import heapq


from types import GeneratorType
def bootstrap(f, stack=[]):
	def wrappedfunc(*args, **kwargs):
		if stack:
			return f(*args, **kwargs)
		else:
			to = f(*args, **kwargs)
			while True:
				if type(to) is GeneratorType:
					stack.append(to)
					to = next(to)
				else:
					stack.pop()
					if not stack:
						break
					to = stack[-1].send(to)
			return to

	return wrappedfunc


def dfs(g, timeIn, timeOut,depths,parents): 
	# assign In-time to node u 
	cnt = 0
	# node, neig_i, parent, depth
	stack = [[1,0,0,0]]
	while stack:
		v,neig_i,parent,depth = stack[-1]
		parents[v] = parent
		depths[v] = depth
		# print (v)
		if neig_i == 0:
			timeIn[v] = cnt    
			cnt+=1
		while neig_i < len(g[v]):
			u = g[v][neig_i]
			if u == parent:
				neig_i+=1
				continue
			stack[-1][1] = neig_i + 1
			stack.append([u,0,v,depth+1])
			break
		if neig_i == len(g[v]):
			stack.pop()
			timeOut[v] = cnt
			cnt += 1
  
# def isAncestor(u: int, v: int, timeIn: list, timeOut: list) -> str: 
# 	return timeIn[u] <= timeIn[v] and timeOut[v] <= timeOut[u] 

cnt = 0
@bootstrap
def dfs(v,adj,timeIn, timeOut,depths,parents,parent=0,depth=0):
	global cnt
	parents[v] = parent
	depths[v] = depth
	timeIn[v] = cnt
	cnt+=1
	for u in adj[v]:
		if u == parent:
			continue
		yield dfs(u,adj,timeIn,timeOut,depths,parents,v,depth+1)
	timeOut[v] = cnt
	cnt+=1
	yield 


def solution(a,n):
	d = defaultdict(int)
	a = [list(reversed(bi(x).zfill(20))) for x in a]
	for i in range(20):
		for x in a:
			if x[i] == '1':
				d[i]+=1
	sol = [['0']*20 for _ in range(n)]
	# return
	for i in range(20):
		j = 0
		while d[i] > 0:
			sol[j][i]='1'
			j+=1
			d[i]-=1
	sol = [int(''.join(reversed(x)),2) for x in sol]
	return sum(x**2 for x in sol)

	

	
def main():
	T = 1
	# T = ri()
	for i in range(T):
		# s = rs()
		n = ri()
		# a,b,n = ria()
		a=ria()

		x = solution(a,n)
		if 'xrange' not in dir(__builtins__):
			print(x)
		else:
			print >>output,str(x)# "Case #"+str(i+1)+':',
	if 'xrange' in dir(__builtins__):
		print(output.getvalue())
		output.close()

if 'xrange' in dir(__builtins__):
	import cStringIO
	output = cStringIO.StringIO()
#example usage:
#    for l in res:
#       print >>output, str(len(l)) + ' ' +  ' '.join(l)

if __name__ == '__main__':
	main()

