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>

#define rep(i, n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

int main() {
  ll n;
  cin >> n;
  vector<ll> a(n);
  vector<pair<ll, ll>> ans;

  rep(i, n) cin >> a[i];
  sort(a.begin(), a.end());

  for (int i = 1; i < n - 1; ++i) {
    if (a[i] > 0) {
      ans.emplace_back(a[0], a[i]);
      a[0] -= a[i];
    } else {
      ans.emplace_back(a[n - 1], a[i]);
      a[n - 1] -= a[i];
    }
  }

  ans.emplace_back(a[n - 1], a[0]);
  cout << a[n - 1] - a[0] << endl;
  for (auto b : ans) printf("%d %d\n", b.first, b.second);
  return 0;
}
