#include <bits/stdc++.h>
using namespace std;
int k, n;
int main() {
    cin >> n; 
    if(!(n & (n - 1))) {printf("No\n"); return 0;} 
    printf("Yes\n");
    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("%d %d\n", i, 1 + n); printf("%d %d\n", i + 1, 1 + n);
        printf("%d %d\n", i + n, i + 1); printf("%d %d\n", i + n + 1, i);
    }
    if(n % 2 == 0){
        for(int i = 2; i < n; i++){
            int p = i ^ 1 ^ n;
            if(p < n) {
                if(i == 2) printf("%d %d\n", i + n, n);
                else printf("%d %d\n", i, n);
                printf("%d %d\n", p, n + n);
                return 0;
            }
        }
    }
}
### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int k, n;
int main() {
    cin >> n; 
    if(!(n & (n - 1))) {printf("No\n"); return 0;} 
    printf("Yes\n");
    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("%d %d\n", i, 1 + n); printf("%d %d\n", i + 1, 1 + n);
        printf("%d %d\n", i + n, i + 1); printf("%d %d\n", i + n + 1, i);
    }
    if(n % 2 == 0){
        for(int i = 2; i < n; i++){
            int p = i ^ 1 ^ n;
            if(p < n) {
                if(i == 2) printf("%d %d\n", i + n, n);
                else printf("%d %d\n", i, n);
                printf("%d %d\n", p, n + n);
                return 0;
            }
        }
    }
}
```