Given an integer N, find two permutations:  Permutation p of numbers from 1 to N such that p_{i} ≠ i and p_{i} & i = 0 for all i = 1, 2, ..., N.  Permutation q of numbers from 1 to N such that q_{i} ≠ i and q_{i} & i ≠ 0 for all i = 1, 2, ..., N. 

& is the bitwise AND operation.


-----Input-----

The input consists of one line containing a single integer N (1 ≤ N ≤ 10^5).


-----Output-----

For each subtask, if the required permutation doesn't exist, output a single line containing the word "NO"; otherwise output the word "YES" in the first line and N elements of the permutation, separated by spaces, in the second line. If there are several possible permutations in a subtask, output any of them.


-----Examples-----
Input
3

Output
NO
NO

Input
6

Output
YES
6 5 4 3 2 1 
YES
3 6 2 5 1 4

Difficulty: interview

Starter code:

Solutions:
def test(x, i):
	i = list(i)
	ok = True
	for j in range(x):
		if (i[j] == j+1 or (i[j]&(j+1) != 0)):
			ok = False
	if ok:
		print(i)

def comp(n):
	return 2**len(bin(n)[2:])-1-n

n = int(input())
nn = n
if (n%2 == 0):
	x = []
	while (n != 0):
		#add n to comp(n) to the front of x
		for i in range(comp(n), n+1):
			x.append(i)
		n = comp(n)-1

	x.reverse()
	print("YES")
	print(' '.join([str(i) for i in x]))
else:
	print("NO")

pow2 = [2**i for i in range(20)]
def make(n):
	if n <= 5: return []
	if n == 6: return [3, 6, 1, 5, 4, 2]
	if n == 7: return [3, 6, 1, 5, 4, 7, 2]
	if n in pow2:
		return []
	shift = 2**(len(bin(n)[2:])-1)
	array = [i for i in range(shift, n+1)]
	array = array[1:] + [array[0]]
	return make(shift-1) + array

n = nn
k = make(n)
if k == []:
	print("NO")
else:
	print("YES")
	print(' '.join([str(i) for i in k]))


from math import log2, floor
n=int(input())
# n=100
if n%2:
	print('NO')
else:
	print("YES")
	k=floor(log2(n))+1
	res = [0 for i in range(n)]
	occupied = [0 for i in range(n)]
	switch = 0
	for i in range(n,0,-1):
		val = 2**k-1-i
		if val > n:
			k-=1
			val = 2**k-1-i
		if occupied[val-1]:
			while occupied[val-1]:
				k-=1
				val = 2**k-1-i
		val = 2**k-1-i
		occupied[val-1] = 1
		res[i-1] = val
	for i in res:
		print(i,end=' ')
	print()

if n in [1,2,3,4,5]:
	print('NO')
elif int(log2(n)) == log2(n):
	print('NO')
elif n == 6:
	print('YES')
	print('3 6 2 5 1 4')
else:
	print('YES')
	print('7 3 2 5 6 4 1',end=' ')
	for i in range(8,n):
		p = pow(2,floor(log2(i))+1)
		if i < pow(2,floor(log2(n))):
			print(p*3//2-1-i,end=' ')
		else:
			print(i+1,end=' ')
	if n > 7:
		print(pow(2,floor(log2(n))))



from math import log2, floor
n=int(input())
# n=100
if n%2:
	print('NO')
else:
	print("YES")
	k=floor(log2(n))+1
	res = [0 for i in range(n)]
	occupied = [0 for i in range(n)]
	switch = 0
	for i in range(n,0,-1):
		val = 2**k-1-i
		if val > n:
			k-=1
			val = 2**k-1-i
		if occupied[val-1]:
			while occupied[val-1]:
				k-=1
				val = 2**k-1-i
		val = 2**k-1-i
		occupied[val-1] = 1
		res[i-1] = val
	for i in res:
		print(i,end=' ')
	print()

if n in [1,2,3,4,5]:
	print('NO')
elif int(log2(n)) == log2(n):
	print('NO')
elif n == 6:
	print('YES')
	print('3 6 2 5 1 4')
else:
	print('YES')
	print('7 3 2 5 6 4 1',end=' ')
	for i in range(8,n+1):
		if int(log2(i)) == log2(i):
			print(min(i*2-1,n),end=' ')
		else:
			print(i-1,end=' ')
	print()



from math import log
n = int(input())
a = [i for i in range(n + 1)]

def ans(n):
    if n <= 0 or n % 2:
        return
    j, cr = (1 << int(log(n, 2))), 1
    while j + cr - 1 <= n:
        a[j - cr], a[j + cr - 1] = a[j + cr - 1], a[j - cr]
        cr += 1
    ans(j - cr)
    
if n % 2 == 0:
    ans(n)
    print("YES")
    print(*a[1:])
else:
    print("NO")

if n <= 5 or (1 << int(log(n, 2))) == n:
    print("NO")
    return
    
print("YES")
print("3 6 1 5 4 2" if n <= 6 else "3 6 1 5 4 7 2", end=' ')
cr = 8
v = (1 << int(log(n, 2)))
for i in range(8, n + 1):
    if i >= v:
        print(n if i == v else i - 1, end=' ')
        continue
    if i == cr:
        cr *= 2
        print(cr - 1, end=' ')
    else:
        print(i - 1, end=' ')

    


from math import*
n=int(input())
if n%2:
	print('NO')
else:
	print("YES")
	k=floor(log2(n))+1
	res = [0 for i in range(n)]
	occupied = [0 for i in range(n)]
	switch = 0
	for i in range(n,0,-1):
		val = 2**k-1-i
		if val > n:
			k-=1
			val = 2**k-1-i
		if occupied[val-1]:
			while occupied[val-1]:
				k-=1
				val = 2**k-1-i
		val = 2**k-1-i
		occupied[val-1] = 1
		res[i-1] = val
	for i in res:
		print(i,end=' ')
	print()

if n in [1,2,3,4,5]:
	print('NO')
elif int(log2(n)) == log2(n):
	print('NO')
elif n == 6:
	print('YES')
	print('3 6 2 5 1 4')
else:
	print('YES')
	print('7 3 2 5 6 4 1',end=' ')
	for i in range(8,n+1):
		if int(log2(i)) == log2(i):
			print(min(i*2-1,n),end=' ')
		else:
			print(i-1,end=' ')
	print()


from math import log2
n = int(input())
p = [0] * (n + 1)
def f(n):
    k = int(log2(n))
    while k:
        i = 1
        while (1 << k) + i - 1 <= n:
            p[(1 << k) + i - 1] = (1 << k) - i
            p[(1 << k) - i] = (1 << k) + i - 1
            i += 1
        n = (1 << k) - i
        if not n:
            break
        k = int(log2(n))
if n & 1:
    print('NO')
else:
    print('YES')
    f(n)
    print(*p[1:])

if n < 6 or 1 << int(log2(n)) == n:
    print('NO')
else:
    print('YES')
    if n == 6:
        print('3 6 2 5 1 4')
    elif n == 7:
        print('7 3 6 5 1 2 4')
    else:
        print('7 3 6 5 1 2 4', end=' ')
        k = 3
        while (1 << k) < n:
            for i in range((1 << k) + 1, min(1 << k + 1, n + 1)):
                print(i, end=' ')
            print(1 << k, end=' ')
            k += 1