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;
#define FOR(i, begin, end) for(ll i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(ll i=(begin)-1;i>=(end);i--)
#define IREP(i, n) IFOR(i,n,0)
#define SORT(a) sort(a.begin(), a.end())
#define ISORT(a) sort(a.begin(), a.end(),greater<ll>())
#define REVERSE(a) reverse(a.begin(), a.end())
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=(" <<; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << ")" << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;}
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)

#define ll long long
#define f first
#define s second
#define mp make_pair
#define pb push_back
#define INF 1000000000000000000

using vec = vector<ll>;
using mat = vector<vec>;
using Pii = pair<ll, ll>;
using PiP = pair<ll, Pii>;
using PPi = pair<Pii, ll>;
using bvec = vector<bool>;
using Pvec = vector<Pii>;

ll mod = 1000000007;

int main()
{
	// 入力
	ll N;
	cin >> N;
	vec Avec(N);
	multiset<ll> A;
	REP(i, N) {
		cin >> Avec[i];
		A.insert(Avec[i]);
	}

	ISORT(Avec);

	ll M = 0;
	if (N % 2 == 0) {
		REP(i, N) {
			if (i < N / 2) M += Avec[i];
			else M -= Avec[i];
		}
		cout << M << endl;

		REP(i, N - 1) {
			auto beginitr = A.begin();
			auto enditr = A.end();
			--enditr;
			if (i % 2 == 1) {
				cout << *beginitr << " " << *enditr << endl;
				ll newA = *beginitr - *enditr;
				A.erase(enditr);
				A.erase(beginitr);
				A.insert(newA);
			}
			else {
				cout << *enditr << " " << *beginitr << endl;
				ll newA = *enditr - *beginitr;
				A.erase(enditr);
				A.erase(beginitr);
				A.insert(newA);
			}

		}
	} else {
		REP(i, N) {
			if (i < (N + 1) / 2) M += Avec[i];
			else M -= Avec[i];
		}
		cout << M << endl;

		REP(i, N - 1) {
			auto beginitr = A.begin();
			auto enditr = A.end();
			--enditr;
			if (i % 2 == 0) {
				cout << *beginitr << " " << *enditr << endl;
				ll newA = *beginitr - *enditr;
				A.erase(enditr);
				A.erase(beginitr);
				A.insert(newA);
			}
			else {
				cout << *enditr << " " << *beginitr << endl;
				ll newA = *enditr - *beginitr;
				A.erase(enditr);
				A.erase(beginitr);
				A.insert(newA);
			}
		}
	}

	return 0;
}