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>

using namespace std;


void solve() {
    int n; cin >> n;
    if (n >= 1 && n == (n&-n)) {
        cout << "No"; return;
    }
    cout << "Yes\n";
    auto p = [](int x, int y){
        cout << x << ' ' << y << "\n";
    };
    p(n+1,n+2); p(n+2, 3); p(3, 1); p(1, 2); p(2, n+3);
    for (int x = 4; x+1 <= n; x+=2) {
        p(x, 1); p(1, x+1);
        p(x, n+x+1); p(x+1, n+x);
    }
    if (n%2 == 0) {
        int x = 1 << (31-__builtin_clz(n));
        int y = n^x;
        y^=1;
        p(x, n); p(y,n+n);
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}