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;
#define rep(i, m, n) for (int i = m; i < n; ++i)

int main() {
  int N; cin >> N;
  vector<int> A(N);
  rep(i, 0, N) cin >> A[i];
  sort(A.begin(), A.end());
  int B = A.back();
  int S = A.front();
  using P = pair<int, int>;
  vector<P> ans;
  rep(i, 1, N-1) {
    if(A[i] >= 0) {
      ans.push_back(P(S, A[i]));
      S -= A[i];
    } else {
      ans.push_back(P(B, A[i]));
      B -= A[i];
    }
  }
  ans.push_back(P(B, S));
  cout << B - S << endl;
  for(P p : ans) cout << p.first << " " << p.second << "\n";
  return 0;
}
