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: Accepted)

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

int main() {
  int n;
  cin>>n;
  int a[n];
  for (int i=0; i<n; i++) cin>>a[i];
  sort(a, a + n);
  queue<int> q1;
  queue<int> q2;
  long long ans = 0;
  if (a[0] >= 0) {
    q2.push(a[0]);
    ans -= a[0];
    for (int i=1; i<n; i++) {
      q1.push(a[i]);
      ans += a[i];
    }
  } else if (a[n-1] <= 0) {
    q1.push(a[n-1]);
    ans += a[n-1];
    for (int i=0; i<n-1; i++) {
      q2.push(a[i]);
      ans -= a[i];
    }
  } else {
    for (int i=0; i<n; i++) {
      if (a[i] >= 0) {
        q1.push(a[i]);
        ans += a[i];
      } else {
        q2.push(a[i]);
        ans -= a[i];
      }
    }
  }
  cout<<ans<<endl;
  while (q1.size() > 0 && q2.size() > 0) {
    int t1 = q1.front();
    int t2 = q2.front();
    q1.pop();
    q2.pop();
    if (q1.size() > 0 && q2.size() == 0) {
      cout<<t2<<" "<<t1<<endl;
      q2.push(t2-t1);
    } else {
      cout<<t1<<" "<<t2<<endl;
      q1.push(t1-t2);
    }
  }
}