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>
using namespace std;
using lint = long long;
using P = pair<int, int>;
using vec = vector<int>;
using mat = vector<vector<int>>;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define endl "\n"

constexpr int MOD = 1000000007;
const int INF = 1 << 30;

int main() {
    int n;
    cin >> n;
    vec a(n);
    rep(i, n) cin >> a[i];

    sort(all(a));

    int x1 = a[n-1], x2 = a[0];

    vector<P> history;

    rep(i, n) {
        if (a[i] > 0 && i > 0 && i < n-1) {
            int y = a[i];
            history.push_back({x2, y});
            x2 -= y;
        }
    }

    rep(i, n) {
        if (a[i] <= 0 && i > 0 && i < n-1) {
            int y = a[i];
            history.push_back({x1, y});
            x1 += abs(y);
        }
    }

    history.push_back({x1, x2});
    x1 -= x2;

    cout << x1 << endl;
    rep(i, n-1) {
        printf("%d %d\n", history[i].first, history[i].second);
    }

    return 0;
}