An undirected graph is called k-regular, if the degrees of all its vertices are equal k. An edge of a connected graph is called a bridge, if after removing it the graph is being split into two connected components.

Build a connected undirected k-regular graph containing at least one bridge, or else state that such graph doesn't exist.


-----Input-----

The single line of the input contains integer k (1 ≤ k ≤ 100) — the required degree of the vertices of the regular graph.


-----Output-----

Print "NO" (without quotes), if such graph doesn't exist. 

Otherwise, print "YES" in the first line and the description of any suitable graph in the next lines.

The description of the made graph must start with numbers n and m — the number of vertices and edges respectively. 

Each of the next m lines must contain two integers, a and b (1 ≤ a, b ≤ n, a ≠ b), that mean that there is an edge connecting the vertices a and b. A graph shouldn't contain multiple edges and edges that lead from a vertex to itself. A graph must be connected, the degrees of all vertices of the graph must be equal k. At least one edge of the graph must be a bridge. You can print the edges of the graph in any order. You can print the ends of each edge in any order.

The constructed graph must contain at most 10^6 vertices and 10^6 edges (it is guaranteed that if at least one graph that meets the requirements exists, then there also exists the graph with at most 10^6 vertices and at most 10^6 edges). 


-----Examples-----
Input
1

Output
YES
2 1
1 2



-----Note-----

In the sample from the statement there is a suitable graph consisting of two vertices, connected by a single edge.

Difficulty: interview

Starter code:

Solutions:
import math,string,itertools,fractions,heapq,collections,re,array,bisect
from itertools import chain, dropwhile, permutations, combinations
from collections import defaultdict, deque

def VI(): return list(map(int,input().split()))

def main1(k):
    # works correctly, but too slow and generates many more vertices than necessary.
    # doesn't pass the time constraint with this implementation. (prints 10^5 lines)
    # ==> use the editorial solution below.
    if k%2==0:
        print("NO")
        return
    print("YES")
    n = 2*(k**2-k+1)
    m = n*k//2
    print(n, m)
    i = 0
    g = [[] for i in range(n+1)]
    print(1,n//2+1)
    off = 1
    for j in range(0,k-1,2):
        j1 = off + j+1
        j2 = off + j+2
        print(off,j1)
        print(off,j2)
        l1 = off + k + j*(k-1)
        l2 = off + k + (j+1)*(k-1)
        for l in range(k-1):
            print(j1, l1+l)
            print(j2, l2+l)
            for m in range(k-1):
                print(l1+l,l2+m)
    off = n//2+1
    for j in range(0,k-1,2):
        j1 = off + j+1
        j2 = off + j+2
        print(off,j1)
        print(off,j2)
        l1 = off + k + j*(k-1)
        l2 = off + k + (j+1)*(k-1)
        for l in range(k-1):
            print(j1, l1+l)
            print(j2, l2+l)
            for m in range(k-1):
                print(l1+l,l2+m)

def main(k):
    # following the editorial algo
    if k%2==0:
        print("NO")
        return
    print("YES")
    if k==1:
        print("2 1")
        print("1 2")
        return
    n = 2*k+4
    m = n*k//2
    e = []
    e.extend([(1,n//2+1)])
    off = 1
    for j in range(off+1,off+k):
        e.extend([(off, j)])
    for j in range(off+1,off+k):
        for i in range(j+1,off+k):
            if (i==j+1 and (j-off)%2==1):# or (j==off+1 and i==off+k-1):
            #if (i==j+1 and i%2==0) or (j==off+1 and i==off+k-1):
                continue
            e.extend([(j,i)])
        e.extend([(j,off+k),(j,off+k+1)])
    e.extend([(off+k,off+k+1)])
    off = n//2+1
    for j in range(off+1,off+k):
        e.extend([(off, j)])
    for j in range(off+1,off+k):
        for i in range(j+1,off+k):
            if (i==j+1 and (j-off)%2==1):# or (j==off+1 and i==off+k-1):
                continue
            e.extend([(j,i)])
        e.extend([(j,off+k),(j,off+k+1)])
    e.extend([(off+k,off+k+1)])
    print(n, m)
    for x in e:
        print(*x)



def main_input(info=0):
    k = int(input())
    main(k)

def __starting_point():
    main_input()

__starting_point()