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: Runtime Error)

#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define pi 3.141592653589793
using namespace std;


int main() {
//    cout << setprecision(10);
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    int n;
    cin >> n;
    ll a[n];
    ll sum = 0;
    vector<ll> pos, neg;
    int z = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        if (a[i] > 0) pos.pb(a[i]);
        else if (a[i] < 0) neg.pb(a[i]);
        else z++;
        sum += abs(a[i]);
    }


    if (pos.empty()) {
        if (z) neg.pb(0);
        sort(neg.begin(), neg.end(), greater<ll>());
        ll p = neg[0];
        cout << sum + 2 * neg[0] << '\n';
        for (int i = 1; i < n; i++) {
            cout << p << ' ' << neg[i] << '\n';
            p -= neg[i];
        }

        return 0;
    }

    if (neg.empty()) {
        if (z) pos.pb(0);
        sort(pos.begin(), pos.end());
        cout << sum - 2 * pos[0] << '\n';
        ll p = pos[0];
        for (int i = 1; i < n - 1; i++) {
            cout << p << ' ' << pos[i] << '\n';
            p -= pos[i];
        }

        cout << pos.back() << ' ' << p;
        return 0;
    }


    if (n == 2 && pos.size() == 1 && neg.size() == 1) {
        cout << pos[0] - neg[0] << '\n';
        cout << pos[0] << ' ' << neg[0];
        return 0;
    }


    cout << sum << '\n';
    ll p = neg.front();
    for (int i = 0; i < pos.size() - 1; i++) {
        cout << p << ' ' << pos[i] << '\n';
        p -= pos[i];
    }

    neg.pb(p);
    p = pos.back();
    for (int i = 1; i < neg.size(); i++) {
        cout << p << ' ' << neg[i] << '\n';
        p -= neg[i];
    }

}