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;

int lowbit(int x) { return x & -x; }

int N;
vector<pair<int, int>> edges;

int main() {
    scanf("%d", &N);
    if (N == lowbit(N)) {
        printf("No\n");
        return 0;
    }
    printf("Yes\n");
    edges.emplace_back(1, 3);
    edges.emplace_back(3, 2);
    edges.emplace_back(1, N + 2);
    edges.emplace_back(N + 2, N + 3);
    edges.emplace_back(N + 3, N + 1);
    for (int i = 4; i < N; i += 2) {
        edges.emplace_back(1, i + 1);
        edges.emplace_back(i + 1, i);
        edges.emplace_back(1, N + i);
        edges.emplace_back(N + i, N + i + 1);
    }
    if (!(N & 1)) {
        int t = lowbit(N);
        edges.emplace_back(N + t, N);
        edges.emplace_back(N - t + 1, N + N);
    }
    for (const auto &i : edges) printf("%d %d\n", i.first, i.second);
}