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 CPP):

Solution:
#include <cstdio>
#include <algorithm>
#include <cctype>
#include <cstring>
using namespace std;
typedef long long i64;
inline int read(int f = 1, int x = 0, char ch = ' ')
{
    while(!isdigit(ch = getchar())) if(ch == '-') f = -1;
    while(isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    return f*x;
}
const int N = 1e3+5;
int k, n, a[N][N];
int main()
{
    n = read(); if((n&-n) == n) return puts("No"), 0;
    puts("Yes");
    printf("1 2\n2 3\n3 %d\n%d %d\n%d %d\n", n+1, n+1, n+2, n+2, n+3);
    for(int i = 2; i*2+1 <= n; ++i) 
        printf("1 %d\n1 %d\n%d %d\n%d %d\n", 2*i, 2*i+1, 2*i, n+2*i+1, 2*i+1, n+2*i);
    if(~n&1)
        for(int x = 2; x <= n; ++x)
        {
            int y = n^x^1; if(x == 3||y == 3||y > n) continue;
            printf("%d %d\n%d %d\n", n, x, n*2, y); break;
        }
    return 0;
}