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 <bits/stdc++.h>

using namespace std;

int main() {
  int n;
  scanf("%d", &n);
  vector<int> vx;
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    vx.push_back(x);
  }
  sort(vx.begin(), vx.end());
  int kecil = vx[0], besar = vx.back();
  vector<pair<int, int>> ans;
  for (int i = 1; i + 1 < n; ++i) {
    int x = vx[i];
    if (x >= 0)
      ans.emplace_back(kecil, x), kecil -= x;
    else
      ans.emplace_back(besar, x), besar -= x;
  }
  ans.emplace_back(besar, kecil);
  printf("%d\n", besar - kecil);
  for (auto it : ans)
    printf("%d %d\n", it.first, it.second);
  return 0;
}
