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, x, y) for (int i = x; i < y; i++)

const int8_t  INFC = 0x3f;
const int16_t INFS = 0x3f3f;
const int32_t INF  = 0x3f3f3f3f;
const int64_t INFL = 0x3f3f3f3f3f3f3f3fLL;
const int     MAXN = 2e5;

int64_t a[ MAXN ];

int main()
{
    ios::sync_with_stdio(0);
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) { cin >> a[ i ]; }
    sort(a, a + n);
    if (n == 2)
    {
        cout << a[ 1 ] - a[ 0 ] << endl << a[ 1 ] << " " << a[ 0 ] << endl;
        return 0;
    }
    int nonPositive = 0;
    for (int i = 0; i < n; i++)
    {
        if (a[ i ] <= 0) nonPositive++;
    }
    int64_t ans=-a[0];
    for (int i=1;i<n;i++)
    {
        ans+=abs(a[i]);
    }
    ostringstream sout;
    nonPositive=max(1,nonPositive);
    nonPositive=min(n-1,nonPositive);
    for (int i=nonPositive;i<n-1;i++)
    {
        sout <<a[0] << " "<<a[i]<< endl;
        a[0]-=a[i];
    }
    for (int i=0;i<nonPositive;i++)
    {
        sout <<a[n-1] << " "<<a[i]<< endl;
        a[n-1]-=a[i];
    }
    cout << a[n-1] << endl;
    cout<<sout.str();
    return 0;
}