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}

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}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
#include <algorithm>
#include <iostream>

using namespace std;

int main() {
  int n; cin >> n;
  int a[100000], p = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (a[i] > 0) ++p;
  }
  
  sort(a, a+n);
  
  if (p == 0) ++p;
  if (p == n) --p;
  
  int m = 0;
  for (int i = 0; i < n-p; ++i) m -= a[i];
  for (int i = n-p; i < n; ++i) m += a[i];
  
  cout << m << endl;
  
  int sm1 = a[n-1];
  for (int i = 1; i < n-p; ++i) {
    cout << sm1 << " " << a[i] << endl;
    sm1 -= a[i];
  }
  
  int sm2 = a[0];
  for (int i = n-p; i < n-1; ++i) {
    cout << sm2 << " " << a[i] << endl;
    sm2 -= a[i];
  }
  
  cout << sm1 << " " << sm2 << endl;
  return 0;
}