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>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
using namespace std;
typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
#define int long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define all(x) (x).begin(), (x).end()
#define forn(i,n) for(int i=0;i<(int)(n);i++)
#define for1(i,n) for(int i=1;i<=(int)(n);i++)
const int MOD = 1e9 + 7 ;
int mpow(int a,int b,int p=MOD){a=a%p;int res=1;while(b>0){if(b&1)res=(res*a)%p;a=(a*a)%p;b=b>>1;}return res%p;}
const int N=400002;

int32_t main(){
      ios_base::sync_with_stdio(false);
      cin.tie(NULL);
      int n;
      cin>>n;
      int a[n];
      bool f1=false,f2=false;
      for(int i=0;i<n;i++) cin>>a[i];
      sort(a,a+n);
      for(int i=0;i<n;i++) if(a[i]<0) f1=true; else f2=true;
      if(f1&&f2)
      {
            vector<int> neg;
            vector<int> pos;
            int sum=0;
            for(int i=0;i<n;i++)
            {
                  if(a[i]>=0) pos.pb(a[i]);
                  else neg.pb(a[i]);
                  sum+=abs(a[i]);
            }
            vector<pii> ans;
            int o=neg.back();
            neg.pop_back();
            for(int i=0;i<(int)pos.size()-1;i++)
            {
                  ans.pb({o,pos[i]});
                  o=o-pos[i];
            }
            ans.pb({pos[(int)pos.size()-1],o});
            o=pos[(int)pos.size()-1]-o;
            for(int i=0;i<neg.size();i++)
            {
                  ans.pb({o,neg[i]});
                  o-=neg[i];
            }
            cout<<o<<"\n";
            for(auto x:ans) cout<<x.fi<<" "<<x.se<<"\n";
      }
      else{
            if(f1)
            {
                  int o=a[0];
                  vector<pii> ans;
                  if(n==2)
                  {
                        ans.pb({a[1],o});
                        o=a[1]-o;
                  }
                  else{
                        ans.pb({o,a[1]});
                        o-=a[1];
                        for(int i=2;i<n;i++)
                        {
                              ans.pb({o,a[i]});
                              o-=a[i];
                        }
                  }
                  cout<<o<<"\n";
                  for(auto x:ans) cout<<x.fi<<" "<<x.se<<"\n";
            }
            else{
                  int o=a[0];
                  vector<pii> ans;
                  if(n==2)
                  {
                        ans.pb({a[1],o});
                        o=a[1]-o;
                  }
                  else{
                        ans.pb({o,a[1]});
                        o-=a[1];
                        for(int i=2;i<n-1;i++)
                        {
                              ans.pb({o,a[i]});
                              o-=a[i];
                        }
                  }
                  cout<<o<<"\n";
                  for(auto x:ans) cout<<x.fi<<" "<<x.se<<"\n";     
            }
      }
}
// I never lose. I either win or I learn