Problem:
Score : 500 points

Problem Statement

There are N integers, A_1, A_2, ..., A_N, written on a blackboard.

We will repeat the following operation N-1 times so that we have only one integer on the blackboard.

Choose two integers x and y on the blackboard and erase these two integers. Then, write a new integer x-y.

Find the maximum possible value of the final integer on the blackboard and a sequence of operations that maximizes the final integer.

Constraints

2 \leq N \leq 10^5

-10^4 \leq A_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible value M of the final integer on the blackboard, and a sequence of operations x_i, y_i that maximizes the final integer, in the format below.

Here x_i and y_i represent the integers x and y chosen in the i-th operation, respectively.

If there are multiple sequences of operations that maximize the final integer, any of them will be accepted.

M
x_1 y_1
:
x_{N-1} y_{N-1}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.*;

 class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        Integer[] x = new Integer[N];
        for(int i = 0; i<N; i++) {
        	x[i] = sc.nextInt();
        }
        Arrays.sort(x, Comparator.reverseOrder());
//        for(int i = 0; i<N; i++) {
//        	System.out.println(x[i]);
//        }

		int tmp=0;
        if(N%2==0) {
        	for(int j=0;j<N-1;j++) {
        		// A最大-最小
        		if(j==0) {
            		System.out.println(x[j] + " " + x[N-(j+1)]);
            		tmp = x[j]-x[N-(j+1)];
            		System.out.println("最初tmp= " + tmp);
        		}
        		else if(j%2==1) {
        			System.out.println(x[N-(j/2+2)] + " " + tmp);
        			tmp = x[N-(j/2+2)] - tmp;
            		System.out.println("奇数tmp= " + tmp);
        		} else {
        			System.out.println(x[j/2] + " " + tmp);
        			tmp = x[j/2] - tmp;
            		System.out.println("偶数tmp= " + tmp);
        		}	
        	}
        }
        else {
        	for(int j=0;j<N-2;j++) {
        		// A最大-最小
        		if(j==0) {
            		System.out.println(x[N-(j+1)] + " " + x[j+1]);
            		tmp = x[N-(j+1)] - x[j+1];
//            		System.out.println("最初tmp= " + tmp);
        		}
        		else if(j%2==0) {
        			System.out.println(x[N-(j/2+1)] + " " + tmp);
        			tmp = x[N-(j/2+1)] - tmp;
//            		System.out.println("奇数tmp= " + tmp);
        		} else {
        			System.out.println(x[j/2 + 2] + " " + tmp);
        			tmp = x[j/2 + 2] - tmp;
 //           		System.out.println("偶数tmp= " + tmp);
        		}	
        	}
			System.out.println(x[0] + " " + tmp);
        	tmp = x[0] - tmp;
        }
        
        
        
        
        int ans=tmp;
        if(N>1) {          	
         ans = tmp;	
        }
        System.out.println(ans);
    }
}