#include <bits/stdc++.h>

using namespace std;

const int max_n = 100111, inf = 1000111222;

int n;

void write(int x, int to) {
    if (x % 2) {
        x += n;
    }
    printf("%d %d\n", x, to);
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    scanf("%d", &n);
    int x = 1;
    while (2 * x <= n) {
        x *= 2;
    }
    if (x == n) {
        puts("No");
        return 0;
    }
    puts("Yes");
    for (int i = 2; i < n; i += 2) {
        printf("1 %d\n", i);
        printf("%d %d\n", i, i + 1);
        printf("1 %d\n", n + i + 1);
        printf("%d %d\n", n + i + 1, n + i);
    }
    printf("3 %d\n", n + 1);
    if (n % 2 == 0) {
        int y = n ^ x ^ 1;
        write(x, n);
        write(y, 2 * n);
    }
    return 0;
}

### Prompt

Your task is to create 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;

const int max_n = 100111, inf = 1000111222;

int n;

void write(int x, int to) {
    if (x % 2) {
        x += n;
    }
    printf("%d %d\n", x, to);
}

int main() {
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    scanf("%d", &n);
    int x = 1;
    while (2 * x <= n) {
        x *= 2;
    }
    if (x == n) {
        puts("No");
        return 0;
    }
    puts("Yes");
    for (int i = 2; i < n; i += 2) {
        printf("1 %d\n", i);
        printf("%d %d\n", i, i + 1);
        printf("1 %d\n", n + i + 1);
        printf("%d %d\n", n + i + 1, n + i);
    }
    printf("3 %d\n", n + 1);
    if (n % 2 == 0) {
        int y = n ^ x ^ 1;
        write(x, n);
        write(y, 2 * n);
    }
    return 0;
}

```