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 <bits/stdc++.h>

const int N = 100010;

int main(){
    int n;
    scanf("%d", &n);
    int x = n & (-n);
    if (x == n){
        puts("No");
        return 0;
    }
    puts("Yes");
    for (int i = 2; i + 1 <= n; i += 2){
        printf("%d %d\n", i + 1, i);
        printf("%d %d\n", i, 1);
        printf("%d %d\n", 1, i + 1 + n);
        printf("%d %d\n", i + 1 + n, i + n);
    }
    printf("%d %d\n", 2 + n, 1 + n);
    if (n % 2 == 0){
        printf("%d %d\n", n, x);
        printf("%d %d\n", n + n, (n ^ x ^ 1) + n);
    }
    return 0;
}