Problem: An undirected graph is called k-regular, if the degrees of all its vertices are equal k. An edge of a connected graph is called a bridge, if after removing it the graph is being split into two connected components.

Build a connected undirected k-regular graph containing at least one bridge, or else state that such graph doesn't exist.

Input

The single line of the input contains integer k (1 ≤ k ≤ 100) — the required degree of the vertices of the regular graph.

Output

Print "NO" (without quotes), if such graph doesn't exist. 

Otherwise, print "YES" in the first line and the description of any suitable graph in the next lines.

The description of the made graph must start with numbers n and m — the number of vertices and edges respectively. 

Each of the next m lines must contain two integers, a and b (1 ≤ a, b ≤ n, a ≠ b), that mean that there is an edge connecting the vertices a and b. A graph shouldn't contain multiple edges and edges that lead from a vertex to itself. A graph must be connected, the degrees of all vertices of the graph must be equal k. At least one edge of the graph must be a bridge. You can print the edges of the graph in any order. You can print the ends of each edge in any order.

The constructed graph must contain at most 106 vertices and 106 edges (it is guaranteed that if at least one graph that meets the requirements exists, then there also exists the graph with at most 106 vertices and at most 106 edges). 

Examples

Input

1


Output

YES
2 1
1 2

Note

In the sample from the statement there is a suitable graph consisting of two vertices, connected by a single edge.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Main {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Main().run(in, out);
        out.close();
    }

    int K;
    void run(FastScanner in, PrintWriter out) {
        K = in.nextInt();

        // for graph of even k,
        // if we remove the bridge, we are left with two connected components
        // consider one of the connected components
        // each node in the graph will have 2s edges, except for 1 node which will have
        // 2s-1 edges, for some value of s
        // this is impossible by the handshaking lemma
        if (K %2 == 0) {
            out.println("NO");
            return;
        } else if (K == 1) {
            out.println("YES");
            out.println("2 1");
            out.println("1 2");
            return;
        }

        // form two connected components and connect them with a bridge
        adj = new LinkedList[2*K+10];
        for (int i = 0; i < adj.length; i++) adj[i] = new LinkedList<>();

        int bridge1 = f(1);
        int bridge2 = f(bridge1+1);
        adj[bridge1].add(bridge2);
        numEdges++;

        out.println("YES");
        out.println(bridge2 + " " + numEdges);
        for (int u = 1; u < adj.length; u++) {
            for (int v : adj[u]) out.println(u + " " + v);

        }
        out.println();
    }

    LinkedList<Integer>[] adj;
    int numEdges = 0;
    int f(int startNode) {
        // to have a k connected graph, use k+1 nodes
        int endNode = startNode + K;

        for (int u = startNode; u <= endNode; u++) {
            for (int v = u+1; v <= endNode; v++) {
                adj[u].add(v);
                numEdges++;
            }
        }

        int bridgeNode = endNode+1;

        for (int i = 0; i < K/2; i++) {

            int u = startNode + (2*i);
            int v = adj[u].pollFirst();

            adj[bridgeNode].add(u);
            adj[bridgeNode].add(v);
            numEdges++;
        }
        return bridgeNode;
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }
    }
}