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>
using namespace std;
typedef long long ll;
#define pb pushback
#define fr(i,n) for(int i=0;i<n;i++)
#define ifr(i,n) for(int i=n-1;i>=0;i--)


int main() {
    int n;
	cin >> n;
    int a[n];
    int m=0;
    int x,y;
    fr(i,n){
     cin >> a[i];
     m+=abs(a[i]); 
    }
    sort(a,a+n);
    if(a[0]<0&&0<a[n-1]){
     cout <<m<< endl;
     int plus;
     fr(i,n){
       if(a[i]>0){
         plus = i;
         break;
       }
     }
     x = a[plus-1];
     y = a[plus];
     fr(i,n-1-plus){
       cout <<x<<" "<<y<< endl;
       x = x-y;
       y = a[plus+i+1];
     }
     x += y;
     y = x-y;
     x -= y;
     
     fr(i,plus){
       cout <<x<<" "<<y<< endl;
       x = x-y;
       y = a[i];
     }
    }else if(a[0]>=0){
      cout <<m-2*a[0]<< endl;
      x=a[0];
      y = a[1];
      fr(i,n-2){
       cout <<x<<" "<<y<< endl;
       x = x-y;
       y = a[i+2];
     }
     cout <<y<<" "<<x<< endl;
    }else{
      cout <<m+2*a[0]<< endl;
      x=a[n-1];
      y=a[0];
      fr(i,n-1){
       cout <<x<<" "<<y<< endl;
       x = x-y;
       y = a[i+1];
      }
   }
}