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:
// C.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 1e5 + 200;

int n;

int main()
{
    scanf("%d", &n);
    if (__builtin_popcount(n) == 1)
        puts("No"), exit(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 = 4; i < n; i += 2)
        printf("1 %d\n%d %d\n1 %d\n%d %d\n", i, i, i + 1 + n, i + 1, i + 1, i + n);
    if (!(n & 1))
    {
        for (int i = 2; i <= n; i++)
        {
            int u = i ^ n ^ 1;
            if (i != 3 && u != 3 && u <= n)
            {
                printf("%d %d\n%d %d\n", i, n, u, 2 * n);
                break;
            }
        }
    }
    return 0;
}