You are given an integer N. Determine if there exists a tree with 2N vertices numbered 1 to 2N satisfying the following condition, and show one such tree if the answer is yes. Assume that, for each integer i between 1 and N (inclusive), Vertex i and N+i have the weight i. Then, for each integer i between 1 and N, the bitwise XOR of the weights of the vertices on the path between Vertex i and N+i (including themselves) is i.

Constraints:
- N is an integer.
- 1 <= N <= 10^5

Input:
Input is given from Standard Input in the following format:
N

Output:
If there exists a tree satisfying the condition in the statement, print "Yes"; otherwise, print "No". Then, if such a tree exists, print the 2N-1 edges of such a tree in the subsequent 2N-1 lines. Each pair (a_i, b_i) means that there is an edge connecting Vertex a_i and b_i. The edges may be printed in any order.

Sample Input 1:
3

Sample Output 1:
Yes
1 2
2 3
3 4
4 5
5 6

The sample output represents the following graph.

Sample Input 2:
1

Sample Output 2:
No

# Buggy version:
n = int(input())
nn = n
f = lambda x, y: x + y * nn

# 2の累乗はむり
b = bin(n)
if b.count('0') == len(b) - 2:
    print(-1)
    exit(0)

ret = [
    [f(1, 0), f(2, 0)],
    [f(2, 0), f(3, 0)],
    [f(3, 0), f(1, 1)],
    [f(1, 1), f(2, 1)],
    [f(2, 1), f(3, 1)],
]

if n % 2 == 0:
    m = n ^ 1
    if m % 2 == 0:
        m += n
    ret.append([f(n, 0), f(1, 1)])
    ret.append([f(n, 1), m])
    n -= 1

# nは奇数
while n > 3:
    ret.append([f(n - 1, 0), f(n, 0)])
    ret.append([f(n, 0), f(1, 0)])
    ret.append([f(1, 0), f(n - 1, 1)])
    ret.append([f(n - 1, 1), f(n, 1)])
    n -= 2

print('Yes')
for x, y in ret:
    # x = x if x <= nn else x - nn
    # y = y if y <= nn else y - nn
    print(x, y)


# Fixed version:
n = int(input())
nn = n
f = lambda x, y: x + y * nn
g = lambda x: x if x % 2 == 1 else x + nn

# 2の累乗はむり
b = bin(n)
if b.count('0') == len(b) - 2:
    print('No')
    exit(0)

ret = [
    [f(1, 0), f(2, 0)],
    [f(2, 0), f(3, 0)],
    [f(3, 0), f(1, 1)],
    [f(1, 1), f(2, 1)],
    [f(2, 1), f(3, 1)],
]

if n % 2 == 0:
    p = 2 ** (len(b) - 3)
    q = n ^ 1 ^ p
    ret.append([g(p), f(n, 0)])
    ret.append([g(q), f(n, 1)])
    n -= 1

# nは奇数
while n > 3:
    ret.append([f(n - 1, 0), f(n, 0)])
    ret.append([f(n, 0), f(1, 1)])
    ret.append([f(1, 1), f(n - 1, 1)])
    ret.append([f(n - 1, 1), f(n, 1)])
    n -= 2

print('Yes')
for x, y in ret:
    # x = x if x <= nn else x - nn
    # y = y if y <= nn else y - nn
    print(x, y)

# Buggy version:
n = int(input())
b = 1 << (n.bit_length()-1)
if n - b == 0:
  print("No")
else:
  print("Yes")
  for i in range(1, 4):
    print(i, i+1)
    print(i+n, i+n+1)
  for i in range(4, n, 2):
    print(i, i+1)
    print(i+n, i+n+1)
    print(1, i)
    print(1, i+n+1)
  if (~n)&1:
    print(n, b)
    a = n - b
    if a&1:
      print(n+n, a+n)
    else:
      print(n+n, a)

# Fixed version:
n = int(input())
b = 1 << (n.bit_length()-1)
if n - b == 0:
  print("No")
else:
  print("Yes")
  print(1, 2)
  print(2, 3)
  print(3, n+1)
  print(n+1, n+2)
  print(n+2, n+3)
  for i in range(4, n, 2):
    print(i, i+1)
    print(i+n, i+n+1)
    print(1+n, i+n)
    print(1+n, i+1)
  if (~n)&1:
    print(n, b+n)
    a = n ^ b ^ 1
    print(n+n, a)

# Buggy version:
n = int(input())

if bin(n).count('1') == 1:
    print('No')
else:
    print('1 2')
    print('2 3')
    print('3 {0}'.format(n + 1))
    print('{0} {1}'.format(n + 1, n + 2))
    print('{0} {1}'.format(n + 2, n + 3))

    if n >= 5:
        if n % 2 == 1:
            for i in range(2, n + 1):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
        else:
            for i in range(2, n):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
            print('{0} {1}'.format(n - 1, n))
            print('{0} {1}'.format((n - 1) ^ 1 ^ n, n * 2))


# Fixed version:
n = int(input())

if bin(n).count('1') == 1:
    print('No')
else:
    print('Yes')
    print('1 2')
    print('2 3')
    print('3 {0}'.format(n + 1))
    print('{0} {1}'.format(n + 1, n + 2))
    print('{0} {1}'.format(n + 2, n + 3))

    if n >= 5:
        if n % 2 == 1:
            for i in range(4, n + 1):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
        else:
            for i in range(4, n):
                print('1 {0}'.format(i))
                if i % 2 == 0:
                    print('{0} {1}'.format(i, i + 1 + n))
                else:
                    print('{0} {1}'.format(i, i - 1 + n))
            print('{0} {1}'.format(n - 1, n))
            print('{0} {1}'.format((n - 1) ^ 1 ^ n, n * 2))