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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

int main(){
	int N; cin >> N;
	int A[N], B[N];
	for(int i=0; i<N; i++) cin >> A[i];
	int f=0, f2=0, sum = 0;
	for(int i=0; i<N; i++){
		if(A[i] <= 0) f = 1;
		if(A[i] >= 0) f2 = 1;
		B[i] = A[i];
		sum += A[i];
	}
	int M=0;
	sort(B, B+N);
	if(f == 0){
		M = sum - 2 * B[0];
		cout << M << endl;
		for(int i=0; i<N-1; i++){
			if(i<N-2){
				cout << B[0] << " " << B[i+1] << endl;
				B[0] -= B[i+1];
			}else{
				cout << B[N-1] << " " << B[0] << endl;
			}
		}
	}else if(f2 == 0){
		M = 2 * B[N-1] - sum;
		cout << M << endl;
		for(int i=0; i<N-1; i++){
			cout << B[N-1] << " " << B[i] << endl;
			B[N-1] -= B[i];
		}
	}else{
		int x=0, y=0;
		sum = 0;
		for(int i=0; i<N; i++){
			if(A[i] < 0){ sum -= A[i]; x++; }
			else{ sum += A[i]; y++; }
		}
		cout << sum << endl;
		for(int i=0; i<x-1; i++){
			cout << B[x] << " " << B[i] << endl;
			B[x] -= B[i];
		}
		cout << B[x-1] << " " << B[x] << endl;
		B[x-1] -= B[x];
		for(int i=0; i<y-2; i++){
			cout << B[x-1] << " " << B[x+1+i] << endl;
			B[x-1] -= B[x+1+i];
		}
		cout << B[N-1] << " " << B[x-1] << endl;
	}
	return 0;
}