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 <stdio.h>

int main(void){
    int n; scanf("%d", &n);
    int p = 1; while(p < n) p <<= 1;
    if(p == n){puts("No");}
    else{
        puts("Yes");
        puts("1 2");
        puts("2 3");
        printf("1 %d\n", n+3);
        printf("%d %d\n", n+3, n+2);
        printf("%d %d\n", n+2, n+1);
        int edge = (n % 2 == 0) ? n - 2 : n;
        for(int v = 4; v <= edge; v += 2){
            printf("1 %d\n", v);
            printf("%d %d\n", v, v+1);
            printf("1 %d\n", v+n+1);
            printf("%d %d\n", v+n+1, v+n);
        }
        
        if(n % 2 == 0){
            printf("%d %d\n", n, p >> 1);
            int d = n - (p >> 1) + 1;
            printf("%d %d\n", d + n, 2 * n);
        }
        
    }
    return 0;
}