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;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,P> P1;
typedef pair<P,P> P2;
#define pu push
#define pb push_back
#define mp make_pair
#define eps 1e-7
#define INF 1000000000
#define mod 1000000007
#define fi first
#define sc second
#define rep(i,x) for(long long i=0;i<x;i++)
#define repn(i,x) for(int i=1;i<=x;i++)
#define SORT(x) sort(x.begin(),x.end())
#define ERASE(x) x.erase(unique(x.begin(),x.end()),x.end())
#define POSL(x,v) (lower_bound(x.begin(),x.end(),v)-x.begin())
#define POSU(x,v) (upper_bound(x.begin(),x.end(),v)-x.begin())
vector<pair<string,P> >vec;

void solve(long long N, vector<long long> A){
	vector<long long> a(N-1), b(N-1);
	SORT(A);
	bool rf;
	ll res, l = 0, r = N-1;
	if ((N-1)%2 == 0) {
		res = A[N-1]; 
		rf = false;
		r--;
	} else {
		res = A[0];
		rf = true;
		l++;
	}
	
	for (int i=0; i<N-1; i++) {
		if (rf) {
			a[i] = A[r];
			b[i] = res;
			res = a[i]-b[i]; 
			rf = false;
			r--;
		} else {
			a[i] = A[l];
			b[i] = res;
			res = a[i]-b[i]; 
			rf = true;
			l++;
		}
	}
	cout << res << endl;
	rep(i,N-1) {
		cout << a[i] << " " << b[i] << endl;
	}
}

int main(){	
	long long N;
	scanf("%lld",&N);
	vector<long long> A(N-1+1);
	for(int i = 0 ; i <= N-1 ; i++){
		scanf("%lld",&A[i]);
	}
	solve(N, A);
	return 0;
}