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 ll long long
#define ld long double
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repo(i,n) for(int i = 1; i < (int)(n); i++)
#define pb push_back
#define mp make_pair
#define np next_permutation
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define mod 1000000007
#define pi acos(-1.0)
const ll INF = 1LL<<61;
template<class T>bool chmax(T &a, const T &b) { 
  if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) {
  if (b<a) { a=b; return 1; } return 0; }
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }
ll jou(ll N, ll P){if(P==0) return 1;
  if(P%2==0){ll t = jou(N, P/2);return t*t % mod;
  }return (N * jou(N, P-1)) % mod;}  //jou(n,mod-2)で逆元
//intの最大値2147483647 ≒ 2×10^9
//long longの最大値9223372036854775807 ≒ 9×10^18
//'大文字'+=32;    で小文字に
//  cout << fixed << setprecision (20);   小数点以下2０桁まで
//実行時間制約2秒では２×10^8回くらいまで計算できる






int main() {
  int n;
  cin>>n;

  vector<ll> p(n);
  ll ans=0;
  rep(i,n){
    cin>>p[i];
    ans+=abs(p[i]);
  }

  sort(all(p));

  if(p[0]>0) cout << ans-2*p[0] << endl;
  else if (p[n-1]<0) cout << ans+2*p[n-1] << endl;
  else cout << ans << endl;

  int now=n-2;
  while(p[now]>=0){
    cout << p[0] << " " << p[now] << endl;
    p[0]-=p[now];
    now--;
    if(now==0) break;
  }

  cout << p[n-1] << " " << p[0] << endl;

  p[n-1]-=p[0];

  now=1;
  while(p[now]<0){
    cout << p[n-1] << " " << p[now] << endl;
    p[n-1]-=p[now];
    now++;
    if(now==n-1) break;
  }

}