p02977 AtCoder Grand Contest 035 - Skolem XOR Tree
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
```python

"""

https://atcoder.jp/contests/agc035/tasks/agc035_c

全ての対の頂点に関して、パス上の重みのxorが自分に等しい
1から順に求めていくことを考える

A,Bの関係を考える
木なので、片方のパス上にあればいい→片方にのみ含ませることも、両方に含ませることも可能

1をいい感じに使って構築できないか？

101-100-1-101-100
みたいに、1の位だけ違う数字をまとめて処理すればいい感じになる
もう一つの1は、どっかの葉につけておけばおk
これで、Nが奇数かつ1でない場合は全て構築可能！！

偶数の時は？ 2,4みたいな場合は自明にダメ
6は？
001 010 011 100 101 110
xorで110になるのは010と100

101-100-1-010-011
    |     |
    110   100

→分からんけどとりあえず提出

偶数でも可能な場合がある…
(N+i)-(N+i+1)- -(i)-(i+1)
とする

"""
import sys
N = int(input())

if N == 1:
    print ("No")

elif N % 2 == 1:
    print ("Yes")

    for i in range(2,N+1,2):
        print (1,i)
        print (i,i+1)
        print (1,i+1+N)
        print (i+1+N,i+N)

    print (N+1,3)

else:

    a = None
    b = None

    for i in range(1,N):
        if N ^ i < N:
            a = i
            b = N ^ i

    if a == None:
        print ("No")
        sys.exit()
    
    print ("Yes")
    for i in range(2,N,2):
        print (i,i+1)
        print (N+i,N+i+1)

        #1につけていく
        if i not in (a,b) and i+1 not in (a,b):
            print (1,i)
            print (1,N+i+1)

    pa = a^1
    pb = b^1

    #print ("===")

    print (N+1,a)
    print (a,b)
    print (N,a)
    print (b,2*N)
    print (1,pa)

    print (1,a+N)
    print (pa+N,pb+N)
    
    
    
        
        

```