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 <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <string>


using namespace std;

int pow2(int n) {
    int ans = 1;
    while (n > 0) {
        if (n % 2) return ans;
        n /= 2;
        ans *= 2;
    }
    return 0;
}

int main() {
    int n;
    cin >> n;
    if (n < 3 || pow2(n) == n) {
        cout << "No" << '\n';
        return 0;
    }
    cout << "Yes" << '\n';
    cout << 1+n << ' ' << 2+n << '\n';
    for (int i=1; i<=(n-1)/2; i++) {
        cout << 1     << ' ' << 2*i << '\n';
        cout << 2*i   << ' ' << 2*i+1 << '\n';
        cout << 1     << ' ' << 2*i+1+n << '\n';
        cout << 2*i+n << ' ' << 2*i+1+n << '\n';
    }
    if (n % 2 == 0) {
        cout << pow2(n) * 2 - 2 << ' ' << n << '\n';
        cout << 2*n-1 << ' ' << 2*n << '\n';
    }
    return 0;
}