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;

// types
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
typedef vector<ll> vll;
typedef vector<ld> vld;
typedef vector<pll> vpll;

// macros
#define ALL(a) a.begin(),a.end()
#define SZ(a) ((int)a.size())
#define FI first
#define SE second
#define REP(i,n) for(int i=0;i<((int)n);i++)
#define REP1(i,n) for(int i=1;i<((int)n);i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define PB push_back
#define EB emplace_back
#define MP(a,b) make_pair(a,b)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
#define Decimal fixed<<setprecision(20)
#define INF 1000000000
#define LLINF 1000000000000000000LL

// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double eps = 1e-10;
const int MOD = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);

  ll n;
  cin >> n;
  vll a(n);
  REP(i, n)
    cin >> a[i];

  sort(a.begin(), a.end());
  auto it = lower_bound(a.begin(), a.end(), 0);
  ll lim = it-a.begin();

  ll ans = -a[0];
  REP1(i, lim)
    ans -= a[i];
  for(ll i = lim; i<n; i++)
    ans += a[i];

  cout << ans << endl;
  REP1(i, lim){
    cout << a[n-1] << " " << a[i] << endl;
    a[n-1] -= a[i];
  }

  for(ll i = max(1LL,lim); i<n-1; i++){
    cout << a[0] << " " << a[i] << endl;
    a[0] -= a[i];
  }
  cout << a[n-1] << " " << a[0] << endl;

}