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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

    #include <stdio.h>
     
    int main()
    {
    	int i, j, tmp;
    	int A[100000];
    	int N;
    	scanf("%d", &N);
    	for (i = 0; i < N; ++i)scanf("%d", &A[i]);
    	//昇順ソート（番号が大きくなるほど大きい）
    	for (i = 0; i < N; ++i) {
    		for (j = i + 1; j < N; ++j) {
    			if (A[i] > A[j]) {
    				tmp = A[i];
    				A[i] = A[j];
    				A[j] = tmp;
    			}
    		}
    	}
     
    	for (int i = 1; i < N-1; i++) {
    		printf("%d %d\n", A[0], A[i]);
    		A[0] -= A[i];
    	}
    	printf("%d %d\n", A[N-1], A[0]);
    	A[N - 1] -= A[0];
    	printf("%d", A[N-1]);
    }