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 <iostream>
#include <fstream>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <list>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef __int128 lll;
constexpr ll mod = 1000000007;

int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

ll n;
vector<ll> a;

int main() {
    cin >> n;
    a.resize(n);
    for (auto &e:a) cin >> e;
    sort(a.begin(), a.end());
    multiset<ll> pos, neg;
    for (ll i = 1; i < n - 1; i++) {
        if (a[i] >= 0) pos.insert(a[i]);
        else neg.insert(a[i]);
    }
    pos.insert(a[n - 1]);
    neg.insert(a[0]);
    vector<pair<ll, ll>> h;
    while (!neg.empty()) {
        ll p = *(pos.begin()), q = *(neg.begin());
        pos.erase(pos.begin());
        neg.erase(neg.begin());
        if (pos.size() <= neg.size()) {
            pos.insert(p - q);
            h.emplace_back(p, q);
        } else {
            neg.insert(q - p);
            h.emplace_back(q, p);
        }
    }
    cout << *(pos.begin()) << endl;
    for (auto e:h) {
        cout << e.first << ' ' << e.second << endl;
    }
}