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: Runtime Error)

#include "bits/stdc++.h"
using namespace std;

#define REP(i, n) for(ll i = 0;i < n;i++)
#define REP1(i, n) for(ll i = 1;i < n;i++)
#define REPR(i, n) for(ll i = n-1;i >= 0;i--)
#define FOR(i, m, n) for(ll i = m;i < n;i++)

#define VSORT(v) sort(v.begin(), v.end())
#define VRSORT(v) sort(v.rbegin(), v.rend())

#define ll long long
#define ALL(a)  (a).begin(),(a).end()

#define pb(a) push_back(a)
#define MOD 1000000007

ll now,n,k;






int main(){
  ll ans=0;
  cin>>n;
  vector<ll> a(n);
  vector<ll> prs;
  vector<ll> mns;

  REP(i,n){
    cin>>a[i];
  }

sort(a.begin(), a.end());
ll maxans=0LL;
int pnp=-1;
if (a[0]>0){
  //a[0]のみマイナスで後プラス
  maxans=-a[0];
  mns.emplace_back(a[0]);
  REP(i,n-1){
    maxans+=a[i+1];
    prs.emplace_back(a[i+1]);
  }
}else{
  if (a[n-1]<0){
    maxans=a[n-1];
    prs.emplace_back(a[n-1]);
    REP(i,n-1){
      maxans-=a[i];
      mns.emplace_back(a[i]);
    }
    //a[n-1]のみプラスで後マイナス
  }else{
    REP(i,n){
      if (a[i]>=0){
        pnp=i;
        break;
      }
      maxans-=a[i];
      mns.emplace_back(a[i]);
    }
    REP(i,n-pnp){
      maxans+=a[i+pnp];
      prs.emplace_back(a[i+pnp]);
    }
  }
}





//cout<<(maxdup)<<endl;
cout<<maxans<<endl;
ll m1;
ll p1;
ll rs;
if (prs.size()>=2){
  //最初にマイナスにしたいものを一つ選ぶ
  m1=mns.back();
  mns.pop_back();
  //次にプラスにしたいものを一つ選ぶ
  p1=prs.back();
  prs.pop_back();
  cout<<m1<<" "<<p1<<endl;
  rs=m1-p1;
  //次にプラスにしたいものを1このこすよう全部列挙していく
  for(int i=0;;i++){
    if (prs.size()==1){
      break;
    }
    p1=prs.back();
    prs.pop_back();
    cout<<rs<<" "<<p1<<endl;
    rs=rs-p1;
  }
}else{
  rs=mns.back();
  mns.pop_back();
}
//次にプラスをx、としてx-yをする
p1=prs.back();
prs.pop_back();
cout<<p1<<" "<<rs<<endl;
rs=p1-rs;
//マイナスをyに列挙してい
for(int i=0;;i++){
  if (mns.size()==0)break;
  m1=mns.back();
  mns.pop_back();
  cout<<rs<<" "<<m1<<endl;
  rs=rs-m1;
}



return 0;
}