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>
#define x first
#define y second

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
const ll MOD = 1e9+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;

int n;

int main() {
    scanf("%d",&n);
    if (n==(n&-n)) {
        puts("No");
        return 0;
    }
    puts("Yes");
    for (int i=2;i<n;i+=2) {
        printf("%d %d\n%d %d\n%d %d\n%d %d\n",i,1,i+1,1,i+n+1,i,i+1,i+n);
    }
    printf("%d %d\n",1+n,3+n);
    if (n%2==0) {
        int cnt = 1;
        while (cnt<n) cnt*=2;
        cnt/=2;
        printf("%d %d\n%d %d\n",n,cnt,n+n,(n^cnt^1));
    }

    return 0;
}