Problem: 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 \leq N \leq 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, in the following format:


a_{1} b_{1}
\vdots
a_{2N-1} b_{2N-1}


Here 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.

Examples

Input

3


Output

Yes
1 2
2 3
3 4
4 5
5 6


Input

1


Output

No

Solution (in PYTHON3):

Solution:
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)