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 <bits/stdc++.h>

#define rep(i, n) for(int i = 0; i<(n); i++)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
using namespace std;
using ll = long long;

int N;
vector<ll> A;

void solve() {
    vector<ll> plus, minus;
    for (int i = 0; i < N; ++i) {
        if (A[i] >= 0) plus.push_back(A[i]);
        else minus.push_back(A[i]);
    }
    sort(plus.begin(), plus.end(), greater<ll>());
    sort(minus.begin(), minus.end());
    if (minus.empty()) minus.push_back(plus.back()), plus.pop_back();
    if (plus.empty()) plus.push_back(minus.back()), minus.pop_back();

    vector<pair<ll, ll>> res;
    ll cur = minus[0];
    for (int i = 0; i + 1 < plus.size(); ++i) {
        res.push_back({cur, plus[i]});
        cur -= plus[i];
    }
    res.push_back({plus.back(), cur});
    cur = plus.back() - cur;

    for (int i = 1; i < minus.size(); ++i) {
        res.push_back({cur, minus[i]});
        cur -= minus[i];
    }

    cout << cur << endl;
    for (auto p : res) cout << p.first << " " << p.second << endl;
}

int main() {
    cin >> N;
    A.resize(N);
    rep(i, N) cin >> A[i];
    solve();
}