Cowboy Vlad has a birthday today! There are n children who came to the celebration. In order to greet Vlad, the children decided to form a circle around him. Among the children who came, there are both tall and low, so if they stand in a circle arbitrarily, it may turn out, that there is a tall and low child standing next to each other, and it will be difficult for them to hold hands. Therefore, children want to stand in a circle so that the maximum difference between the growth of two neighboring children would be minimal possible.

Formally, let's number children from 1 to n in a circle order, that is, for every i child with number i will stand next to the child with number i+1, also the child with number 1 stands next to the child with number n. Then we will call the discomfort of the circle the maximum absolute difference of heights of the children, who stand next to each other.

Please help children to find out how they should reorder themselves, so that the resulting discomfort is smallest possible.

Input

The first line contains a single integer n (2 ≤ n ≤ 100) — the number of the children who came to the cowboy Vlad's birthday.

The second line contains integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) denoting heights of every child.

Output

Print exactly n integers — heights of the children in the order in which they should stand in a circle. You can start printing a circle with any child.

If there are multiple possible answers, print any of them.

Examples

Input


5
2 1 1 3 2


Output


1 2 3 2 1


Input


3
30 10 20


Output


10 20 30

Note

In the first example, the discomfort of the circle is equal to 1, since the corresponding absolute differences are 1, 1, 1 and 0. Note, that sequences [2, 3, 2, 1, 1] and [3, 2, 1, 1, 2] form the same circles and differ only by the selection of the starting point.

In the second example, the discomfort of the circle is equal to 20, since the absolute difference of 10 and 30 is equal to 20.


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

/**
 *  
 * @author akash_2ehuikg
 * Coding is love <3!
 */

public class C {
    public static void main(String args[]) {
        try {
            int n=nextInt();
            long a[]=new long[n];
            for (int i = 0; i < n; i++) {
                a[i]=nextLong();
            }
            Arrays.sort(a);
            if(n%2!=0){
                long ans[]=new long[n];
                int index=0;
                for (int i = 0; i+1 < n; i+=2) {
                    ans[index]=a[i];
                    ans[n-index-1]=a[i+1];
                    index++;
                }
                ans[n/2]=a[n-1];
                for (int i = 0; i < n; i++) {
                    bw.write(ans[i]+" ");
                }
                bw.write("\n");
                bw.flush();
            }else{
                long ans[]=new long[n];
                int index=0;
                for (int i = 0; i+1 < n; i+=2) {
                    ans[index]=a[i];
                    ans[n-index-1]=a[i+1];
                    index++;
                }
               
                for (int i = 0; i < n; i++) {
                    bw.write(ans[i]+" ");
                }
                bw.write("\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 boolean prime[]=new boolean[10000001];

    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;
    }

//******************************************************************************
//Sieve
static void sieveOfEratosthenes(boolean prime[]){
        int n=10000000;
        for(int i=0;i<n;i++) 
            prime[i] = true; 
          
        for(int p = 2; p*p <=n; p++){ 
            if(prime[p] == true) 
            {
                for(int i = p*p; i <= n; i += p) 
                    prime[i] = false; 
            } 
        } 
    } 

//******************************************************************************
//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;
    }
    //**************************************************************************
}
