Problem: Mahmoud was trying to solve the vertex cover problem on trees. The problem statement is:

Given an undirected tree consisting of n nodes, find the minimum number of vertices that cover all the edges. Formally, we need to find a set of vertices such that for each edge (u, v) that belongs to the tree, either u is in the set, or v is in the set, or both are in the set. Mahmoud has found the following algorithm:

  * Root the tree at node 1. 
  * Count the number of nodes at an even depth. Let it be evenCnt. 
  * Count the number of nodes at an odd depth. Let it be oddCnt. 
  * The answer is the minimum between evenCnt and oddCnt. 



The depth of a node in a tree is the number of edges in the shortest path between this node and the root. The depth of the root is 0.

Ehab told Mahmoud that this algorithm is wrong, but he didn't believe because he had tested his algorithm against many trees and it worked, so Ehab asked you to find 2 trees consisting of n nodes. The algorithm should find an incorrect answer for the first tree and a correct answer for the second one.

Input

The only line contains an integer n (2 ≤ n ≤ 105), the number of nodes in the desired trees.

Output

The output should consist of 2 independent sections, each containing a tree. The algorithm should find an incorrect answer for the tree in the first section and a correct answer for the tree in the second. If a tree doesn't exist for some section, output "-1" (without quotes) for that section only.

If the answer for a section exists, it should contain n - 1 lines, each containing 2 space-separated integers u and v (1 ≤ u, v ≤ n), which means that there's an undirected edge between node u and node v. If the given graph isn't a tree or it doesn't follow the format, you'll receive wrong answer verdict.

If there are multiple answers, you can print any of them.

Examples

Input

2


Output

-1
1 2


Input

8


Output

1 2
1 3
2 4
2 5
3 6
4 7
4 8
1 2
1 3
2 4
2 5
2 6
3 7
6 8

Note

In the first sample, there is only 1 tree with 2 nodes (node 1 connected to node 2). The algorithm will produce a correct answer in it so we printed  - 1 in the first section, but notice that we printed this tree in the second section.

In the second sample:

In the first tree, the algorithm will find an answer with 4 nodes, while there exists an answer with 3 nodes like this: <image> In the second tree, the algorithm will find an answer with 3 nodes which is correct: <image>

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;

/**
 *  
 * @author akashvermaofskt
 * Coding is love <3!
 */

public class C {
    public static void main(String args[]) {
        try {
            int n=nextInt();
            
            if(n<=5){
                bw.write("-1\n");
            }else{
                for (int i = 2; i <=n-2; i++) {
                    bw.write("1 "+i+"\n");
                }
                bw.write("2 "+(n-1)+"\n");
                bw.write("2 "+n+"\n");
            }
            
            for (int i = 1; i <= n-1; i++) {
                bw.write(i+" "+(i+1)+"\n");
            }
            bw.flush();
        } catch (Exception e) {
            e.printStackTrace();
        }
    }


    //TEMPLATE
    
    //******************************************************************************
    // Fast I/O 
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    static StringTokenizer st = null;
    static int mod=1000000007;

    static String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
    
    static int nextInt() {
        return Integer.parseInt(next());
    }
    
    static long nextLong() {
        return Long.parseLong(next());
    }

    static double nextDouble() {
        return Double.parseDouble(next());
    }

    static int b_search(int l,int r,int A[],int key){
        int ans=0;
        l=0;r=1000000000;

        while(l<r){
            //System.out.println(ans);
            ans=(l+r)/2;
            long temp=0;
            for (int i = 0; i < A.length; i++) {
                temp+=(long)Math.ceil((double)A[i]/(double)ans);
            }

            if(temp<=key){
                r=ans;
            }
            else{
                l=ans+1;
                ans=r;
            }
        }

        return ans;
    }

//******************************************************************************
//Modulo and FastExpo
    static long modInverse(long a)
    {
       return power(a, mod - 2,mod);
    }
     
    // To compute x^y under modulo m
    static long power(long x, long y, int p)
    {
        // Initialize result
        long res = 1;     
        // Update x if it is more  
        // than or equal to p
        x = x % p; 
     while (y > 0)
        {
            // If y is odd, multiply x
            // with result
            if((y & 1)==1)
                res = (res * x) % p;
            // y must be even now
            // y = y / 2
            y = y >> 1; 
            x = (x * x) % p; 
        }
        return res;
    }
//******************************************************************************
//GCD
    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
         
        return gcd(b%a, a);
    }

    //******************************************************************************
    // Useful user datatype
    
    static class Pair{
        public long x;
        public long y;

        public Pair(long first, long second){
            this.x = first;
            this.y = second;
            }
        @Override
        public String toString() {
        return "(" + x + "," + y + ")";
        }
    }

    static Comparator csort=new Comparator<Pair>(){
                    public int compare(Pair a, Pair b) {
                        if(a.x < b.x)
                        return -1;
                        else if (a.x > b.x)
                        return 1;
                        else if(a.y < b.y)
                        return -1;
                        else if (a.y > b.y)
                        return 1;
                        else
                        return 0;
                    }
                };

    //******************************************************************************
    //N choose K
    public static long choose(long total, long choose){
        if(total < choose)
            return 0;
        if(choose == 0 || choose == total)
            return 1;
        return choose(total-1,choose-1)+choose(total-1,choose);
    }

    //******************************************************************************
    //Permutations
    // simply prints all permutation - to see how it works
    private static void printPermutations( Comparable[] c ) {
            System.out.println( Arrays.toString( c ) );
            while ( ( c = nextPermutation( c ) ) != null ) {
                    System.out.println( Arrays.toString( c ) );
            }
    }

    // modifies c to next permutation or returns null if such permutation does not exist
    private static Comparable[] nextPermutation( final Comparable[] c ) {
            // 1. finds the largest k, that c[k] < c[k+1]
            int first = getFirst( c );
            if ( first == -1 ) return null; // no greater permutation
            // 2. find last index toSwap, that c[k] < c[toSwap]
            int toSwap = c.length - 1;
            while ( c[ first ].compareTo( c[ toSwap ] ) >= 0 )
                    --toSwap;
            // 3. swap elements with indexes first and last
            swap( c, first++, toSwap );
            // 4. reverse sequence from k+1 to n (inclusive) 
            toSwap = c.length - 1;
            while ( first < toSwap )
                    swap( c, first++, toSwap-- );
            return c;
    }

    // finds the largest k, that c[k] < c[k+1]
    // if no such k exists (there is not greater permutation), return -1
    private static int getFirst( final Comparable[] c ) {
            for ( int i = c.length - 2; i >= 0; --i )
                    if ( c[ i ].compareTo( c[ i + 1 ] ) < 0 )
                            return i;
            return -1;
    }

    // swaps two elements (with indexes i and j) in array 
    private static void swap( final Comparable[] c, final int i, final int j ) {
            final Comparable tmp = c[ i ];
            c[ i ] = c[ j ];
            c[ j ] = tmp;
    }
    //******************************************************************************
    // DSU
    //Maintain an Array A with N elements and Array size for size of each set, intialize A[i]=i and size[i]=1

    static int root(int A[],int i){
        while(A[i]!=i){
            A[i]=A[A[i]];
            i=A[i];
        }
        return i;
    }

    static boolean find(int A[],int a,int b){
        if(root(A,a)==root(A,b))return true;
        else return false;
    }

    static void union(int A[],int size[],int a,int b){
        int ra=root(A,a);
        int rb=root(A,b);

        if(ra==rb)return;

        if(size[ra]<size[rb]){
            A[ra]=A[rb];
            size[rb]+=size[ra];
        }else{
            A[rb]=A[ra];
            size[ra]+=size[rb];
        }
    }

    //**************************************************************************
    //BinarySearch
    /* binary search for 5:
                v-- lower bound
         1 2 3 4 5 5 5 6 7 9
                      ^-- upper bound
        
            binary search for 8
                        v-- lower bound
       1 2 3 4 5 5 5 6 7 9
                        ^-- upper bound
    */
    //For finding greater than 
    static int upper_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid;

        while (first < last) {
            mid =first+(last - first)/2; 

            if (A[mid] <= key)  
                first = mid + 1; 
            else 
                last = mid;
        }

        return first;
    }
    
    //For finding greater than equal to
    static int lower_bound(int A[],int key){
        int first = 0;
        int last = A.length;
        int mid=0;
        
        while (first < last) {
            mid = first + ((last - first) >> 1); 
            if (A[mid] < key)  
                first = mid + 1; 
            else 
                last = mid;
        }
        
        return first;
    }
    //**************************************************************************
}