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>
#define ll long long
const int Maxn=0;
#define mod 0x3f3f3f3f
using namespace std;
int a[100003],b[100003];
int s[100003];
priority_queue<int,vector<int>,greater<int> >q;
int main()
{  int n;
cin>>n;
for(int i=1;i<=n;i++)
    cin>>s[i];
sort(s+1,s+1+n);
 int Min=s[1]-s[n-1];int sum=0;
 a[++sum]=s[1],b[sum]=s[n-1];
 for(int i=n-2;i>=2;i--)
    {
        a[++sum]=Min;
        b[sum]=s[i];
        Min-=s[i];
    }
    a[++sum]=s[n];
    b[sum]=Min;
    cout<<a[sum]-b[sum]<<endl;
    for(int i=1;i<=sum;i++)
        cout<<a[i]<<" "<<b[i]<<endl;
    return 0;
}