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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstring>
#include <climits>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

constexpr ll INF = 1e18;
constexpr ll MOD = 1e9 + 7;
constexpr double PI = 3.14159265358979323846;
constexpr int di[] = {0, 0, 1, -1};
constexpr int dj[] = {1, -1, 0, 0};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, a[112345];

    cin >> n;
    REP(i, n)
    cin >> a[i];

    sort(a, a + n);

    V p, m;

    ll res = 0;

    if (n % 2)
    {
        for (int i = 0; i < n / 2; i++)
        {
            res -= a[i];
            m.push_back(a[i]);
        }
        for (int i = n / 2 + 1; i < n; i++)
        {
            res += a[i];
            p.push_back(a[i]);
        }

        if (a[n / 2] < 0)
        {
            res -= a[n / 2];
            m.push_back(a[n / 2]);
        }
        else
        {
            res += a[n / 2];
            p.push_back(a[n / 2]);
        }
    }
    else
    {
        for (int i = 0; i < n / 2; i++)
        {
            res -= a[i];
            m.push_back(a[i]);
        }
        for (int i = n / 2; i < n; i++)
        {
            res += a[i];
            p.push_back(a[i]);
        }
    }

    cout << res << endl;

    while (true)
    {
        V np, nm;

        for (int i = 0; i < min(p.size(), m.size()); i++)
        {
            if ((i + (m.size() - p.size() == 1)) % 2 || (p.size() == 1 && m.size() == 1))
            {
                cout << p[i] << " " << m[i] << endl;
                np.push_back(p[i] - m[i]);
            }
            else
            {
                cout << m[i] << " " << p[i] << endl;
                nm.push_back(m[i] - p[i]);
            }
        }

        if (p.size() - m.size() == 1)
            np.push_back(p[p.size() - 1]);
        if (m.size() - p.size() == 1)
            nm.push_back(m[m.size() - 1]);

        p = np;
        m = nm;

        if (p.size() + m.size() == 1)
            return 0;
    }

    return 0;
}