This is the easy version of the problem. The difference between the versions is that in the easy version all prices a_i are different. You can make hacks if and only if you solved both versions of the problem.

Today is Sage's birthday, and she will go shopping to buy ice spheres. All n ice spheres are placed in a row and they are numbered from 1 to n from left to right. Each ice sphere has a positive integer price. In this version all prices are different.

An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.

You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of ice spheres in the shop.

The second line contains n different integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the prices of ice spheres.

Output

In the first line print the maximum number of ice spheres that Sage can buy.

In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.

Example

Input


5
1 2 3 4 5


Output


2
3 1 4 2 5 

Note

In the example it's not possible to place ice spheres in any order so that Sage would buy 3 of them. If the ice spheres are placed like this (3, 1, 4, 2, 5), then Sage will buy two spheres: one for 1 and one for 2, because they are cheap.
import java.io.*;
import java.util.*;
import java.lang.Math;

public class Main{
	static BufferedReader reader;
	static StringTokenizer tokenizer;
	static PrintWriter writer;

	static int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	static long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}

	static String nextToken() throws IOException {
		while (tokenizer == null || !tokenizer.hasMoreTokens()) {
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}
	static String nexts() throws IOException {
	  	  tokenizer = new StringTokenizer(reader.readLine());
	   	 String s="";
	   	 while (tokenizer.hasMoreTokens()) {
			s+=tokenizer.nextElement()+" ";
		}
		return s;
	}

	//String str=nextToken();
	//String[] s = str.split("\\s+");

	public static int gcd(int x, int y){
		if (y == 0) return x; else return gcd(y, x % y);
	}

	public static boolean isPrime(int n) 
	    { 
	        // Corner cases 
	        if (n <= 1){
		 return false; }
	        if (n <= 3){
		 return true; }
	      
	        // This is checked so that we can skip  
	        // middle five numbers in below loop 
	        if (n % 2 == 0 || n % 3 == 0){
		 return false; }
	      
	        for (int i = 5; i * i <= n; i = i + 6) {    //Checking 6i+1 & 6i-1
	            if (n % i == 0 || n % (i + 2) == 0) {
	            	return false;  }
	            }
			//O(sqrt(n))
	        return true; 
	    } 

	public static void shuffle(int[] A) {
		for (int i = 0; i < A.length; i++) {
			int j = (int)(i * Math.random());
			int tmp = A[i];
			A[i] = A[j];
			A[j] = tmp;
		}
	}

	 public static long power(int x, long n)
	{
	    long mod = 1000000007;
		if (n == 0) {
			return 1;
		}

		long pow = power(x, n / 2);

		if ((n & 1) == 1) {
			return (x * pow * pow)%mod;
		}
		return (pow * pow)%mod;
	}
	static long power(long x, long y, long p)     //x^y%p
	    { 
  	       long res = 1; 
	       x = x % p; 
  	       while (y > 0) { 
 	 	 if (y % 2 == 1) {
      	        	        res = (res * x) % p; 
		 }
  		 y = y >> 1; // y = y/2 
     	     	 x = (x * x) % p; 
     	         } 
  	         return res; 
    	} 

	 static long ncr(int n, int k)  
   	 { 
           		 long res = 1; 
  		   // Since C(n, k) = C(n, n-k)  
 		       if (k > n - k) { 
     		       k = n - k; 
      		  } 
  		  // Calculate value of [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]  
     		   for (int i = 0; i < k; ++i) { 
        		    res *= (n - i); 
         		   res /= (i + 1); 
     		   } 
  		  return res; 
 	   }

	static int modInverse(int n, int p) 
  	{ 
    	          return (int)power((long)n, (long)(p - 2),(long)p); 
	 } 
	static int nCrModPFermat(int n, int r, int p) 
   	 { 
	        if (r == 0)        // Base case 
	            return 1; 
  	        int[] fac = new int[n + 1];         // Fill factorial array so that we can find all factorial of r, n and n-r 
	        fac[0] = 1; 
	        for (int i = 1; i <= n; i++) 
                                     fac[i] = fac[i - 1] * i % p; 
   	         return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;     //IMPORTANT
    	}	 
	static int lower(int a[], int x)   //Returns rightest index of x in sorted arr else -1
  	  {                                                //If not present returns index of just smaller element
  	            int n=a.length;
    	        int l=0;
                int r=n-1;
                int ans=-1;
                while(l<=r){
                    int m=(r-l)/2+l;
                    if(a[m]<=x){
                        ans=m;
                        l=m+1;
                    }
                    else{
                        r=m-1;
                    }
                }
                 return ans;  
                } 

	static int upper(int a[], int x)     //Returns leftest index of x in sorted arr else n
 	   {                                                  //If not present returns index of just greater element
 	            int n=a.length;
   	            int l=0;
                int r=n-1;
                int ans=n;
                while(l<=r){
                    int m=(r-l)/2+l;
                    if(a[m]>=x){
                        ans=m;
                        r=m-1;
                    }
                    else{
                         l=m+1;
                    }
                }
                 return ans; 
                 } 

	public static long stringtoint(String s){ // Convert bit string to number
		long a = 1;
		long ans = 0;
		StringBuilder sb = new StringBuilder();
		sb.append(s);  
		sb.reverse();
		s=sb.toString();
		for(int i=0;i<s.length();i++){
			ans = ans + a*(s.charAt(i)-'0');
			a = a*2;
		}
		return ans;
	}
 
	String inttostring(long a,long m){        // a is number and m is length of string required
		String res = "";                     // Convert a number in its string representation
		for(int i=0;i<(int)m;i++){
			if((a&(1<<i))==1){
				res+='1';
			}else
				res+='0';
		}
		StringBuilder sb = new StringBuilder();
		sb.append(res);  
		sb.reverse();
		res=sb.toString();
		return res;
    }
	
	 static class R implements Comparable<R>{
	        int x, y;
	        public  R(int x, int y) {
	            this.x = x;
	            this.y = y;
	        }
	        public int compareTo(R o) {
            		return x-o.x;     //Increasing order(Which is usually required)
        		}
	    }   
	// int t=a[i];
	// a[i]=a[j];
	// a[j]=t;
	//double d=Math.sqrt(Math.pow(Math.abs(x2-x1),2)+Math.pow(Math.abs(y2-y1),2));
	public static void main(String[] args) throws IOException {
		reader = new BufferedReader(new InputStreamReader(System.in));
		tokenizer = null;
		writer = new PrintWriter(System.out);
		solve();
		reader.close();
		writer.close();
	}
	
	private static void solve() throws IOException {
	//	int t = nextInt();
	//	while(t-->0){
			//String s= nextToken();
			//long[] a=new long[n];
			//ArrayList<Integer> ar=new ArrayList<Integer>();
			//HashSet<Integer> set=new HashSet<Integer>();
			//HashMap<Integer,String> h=new HashMap<Integer,String>();
			//R[] a1=new R[n];
			//char[] c=nextToken().toCharArray();

			int n = nextInt();
			//long n = nextLong();
			long[] a=new long[n];
			for(int i=0;i<n;i++){
				a[i]=nextLong();
			}
			//shuffle(a);
			Arrays.sort(a);
			int mm=(n-1)/2;
			writer.println(mm);
			long[] b=new long[n];
			int j=0;
			for(int i=1;i<n-1;){
			    b[i]=a[j];
			    j++;
			    i++;
			    i++;
			}
			for(int i=0;i<n;){
			    b[i]=a[j];
			    j++;
			    if(i==n-2){
			        i++;
			        continue;
			    }
			    i++;
			    i++;
			}
			for(int i=0;i<n;i++){
			    writer.print(b[i]+" ");
			}
               //  (int)
            
			
		 }
//	}
}