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:
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

#A
def A():
    n = I()
    a = LI()
    ans = 0
    for i in a:
        ans ^= i
    if not ans:
        print("Yes")
    else:
        print("No")
    return

#B
def B():
    n,m = LI()

    return

#C
def C():
    n = I()
    p = [(1<<i) for i in range(100)]
    if n in p:
        print("No")
        quit()
    if n+1 in p:
        print("Yes")
        for i in range(1,2*n):
            print(i,i+1)
        quit()
    ans = []
    for i in range(1,3):
        ans.append((i,i+1))
    ans.append((3,n+1))
    for i in range(1,3):
        ans.append((i+n,i+n+1))
    u = 1
    d = 1
    for i in range(2,n//2+n%2):
        ans.append((u,2*i))
        ans.append((d,2*i+1))
        ans.append((2*i,2*i+n+1))
        ans.append((2*i+1,2*i+n))
        u = 2*i+n+1
        d = 2*i+n

    if n%2:
        print("Yes")
        for i,j in ans:
            print(i,j)
    else:
        ans.append((n-1,n))
        for i in range(n):
            if p[i]&n:
                break
        ans.append((p[i+1]-2,2*n))
        print("Yes")
        for i,j in ans:
            print(i,j)
    return

#D
def D():
    n = I()

    return

#E
def E():
    n = I()

    return

#F
def F():
    n = I()

    return

#Solve
if __name__ == "__main__":
    C()