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: Accepted)

#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
const int INF = (1<<30);
const ll INFLL = (1ll<<60);
const ll MOD = (ll)(1e9+7);

#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	a = (a<MOD)?a:(a-MOD);
	b = (b<MOD)?b:(b-MOD);
	a += b;
	a = (a<MOD)?a:(a-MOD);
}

ll a[100100];

int main(void){
	int n,i;
	ll m;
	cin >> n;
	for(i=0; i<n; ++i){
		cin >> a[i];
	}
	sort(a,a+n);
	if(a[0]>=0){
		m = -a[0];
		for(i=1; i<n; ++i){
			m += a[i];
		}
		cout << m << endl;
		for(i=1; i<n; ++i){
			if(i==n-1){
				swap(a[0],a[i]);
			}
			cout << a[0] << " " << a[i] << endl;
			a[0] -= a[i];
		}
	}else if(a[n-1]<=0){
		reverse(a,a+n);
		m = a[0];
		for(i=1; i<n; ++i){
			m -= a[i];
		}
		cout << m << endl;
		for(i=1; i<n; ++i){
			cout << a[0] << " " << a[i] << endl;
			a[0] -= a[i];
		}
	}else{
		m = 0ll;
		for(i=0; i<n; ++i){
			m += abs(a[i]);
		}
		cout << m << endl;
		for(i=1; i<(n-1); ++i){
			if(a[i]>0){
				cout << a[0] << " " << a[i] << endl;
				a[0] -= a[i];
			}else{
				cout << a[n-1] << " " << a[i] << endl;
				a[n-1] -= a[i];
			}
		}
		cout << a[n-1] << " " << a[0] << endl;
	}
	return 0;
}