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;
#define REP(i,a) for(int i = 0; i < (a); i++)
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const int MOD = 1e9 + 7;

int main(){
	int n;
    cin >> n;
    int a[n];
    ll sum = 0;
    REP(i,n){
        cin >> a[i];
        sum += abs(a[i]);
    }
    sort(a, a + n);
    int mn = a[0], mx = a[n - 1];
    if(a[0] * a[n - 1] <= 0){
        cout << sum << endl;
    }else if(mn > 0){
        cout << sum - 2 * mn << endl;
    }else{
        cout << sum + 2 * mx << endl;
    }
    for(int i = 1; i < n - 1; i++){
        if(a[i] >= 0){
            cout << mn << " " << a[i] << endl;
            mn -= a[i];
        }else{
            cout << mx << " " << a[i] << endl;
            mx -= a[i];
        }
    }
    cout << mx << " " << mn << endl;
}