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())

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))