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 int long long
#define For(i, a, b) for (int i = (a); i <= static_cast<int>(b); i++)
#define Forr(i, a, b) for (int i = (a); i >= static_cast<int>(b); i--)
#define rep(i, n) For(i, 0, n - 1)
#define repall(i, arr) for (auto& i : (arr))
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define dump(x) cerr << #x << " = " << (x) << '\n'
#define dump2(x, y) \
    cerr << #x << " = " << (x) << " " << #y << " = " << (y) << '\n'

template <typename T>
using pq = priority_queue<T>;
template <typename T>
using pqr = priority_queue<T, vector<T>, greater<T>>;
const int INF = LLONG_MAX / 2;
constexpr int MOD = 1e9 + 7;
using P = pair<int, int>;
using vec = vector<int>;
using mat = vector<vec>;

template <typename T1, typename T2>
ostream& operator<<(ostream& stream, const pair<T1, T2>& p) {
    return stream << p.first << " " << p.second;
}
template <typename T>
void print(const vector<vector<T>> matrix) {
    repall (vec, matrix)
        print(vec);
}
template <typename T>
void print(const vector<T> vec) {
    unsigned int len = vec.size();
    rep (i, len - 1)
        cout << vec[i] << ' ';
    cout << vec[len - 1] << '\n';
}
template <typename Arg>
void print(const Arg arg) {
    cout << arg << '\n';
}
template <typename Head, typename... Args>
void print(const Head head, const Args... args) {
    cout << head << " ";
    print(args...);
}
template <typename T>
T sum_(vector<T> vec, T init = 0) {
    return std::accumulate(all(vec), T(init));
}

void yn(bool tf) { print(tf ? "Yes" : "No"); }
void YN(bool tf) { print(tf ? "YES" : "NO"); }

template <typename T>
void init(vector<T>& v) {
    rep (i, v.size())
        cin >> v[i];
}

template <typename T, typename U>
void init(vector<T>& v, vector<U>& w) {
    assert(v.size() == w.size());
    rep (i, v.size())
        cin >> v[i] >> w[i];
}

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

// -------------------------------------------------------------------
// sort 1 2 3 4
// pqr 1 2 3 must impl >

signed main() {
    cin.tie(0), cout.tie(0), ios::sync_with_stdio(false);
    int n;
    cin >> n;
    vec li(n);
    init(li);

    sort(all(li));
    if (li[0] >= 0) {
        int sum = -li[0];
        For(i, 1, n - 1) sum += li[i];
        print(sum);
        int tmp = li[0];
        for (int idx = 1; idx <= n - 2; idx++) {
            print(tmp, li[idx]);
            tmp = tmp - li[idx];
        }
        print(li[n - 1], tmp);
    } else if (li[n - 1] < 0) {
        int sum = li[n - 1];
        For(i, 1, n - 1) sum += -li[i];
        print(sum);
        int tmp = li[0];
        for (int idx = n - 2; idx >= 1; idx--) {
            print(tmp, li[idx]);
            tmp = tmp - li[idx];
        }
        print(tmp, li[0]);
    } else {
        int sum = 0;
        rep (i, n)
            sum += abs(li[i]);
        print(sum);
        int f = 0, tmp = 0;

        while (li[f + 1] < 0) {
            print(li[n - 1], li[f]);
            li[n - 1] -= li[f];
            f++;
        }
        tmp = li[f];
        for (int idx = f + 1; idx <= n - 2; idx++) {
            print(tmp, li[idx]);
            tmp = tmp - li[idx];
        }
        print(li[n - 1], tmp);
    }

    return 0;
}