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, n) for(int i=0; i<(int)(n); i++)
int main(){
  int N; long A; cin >> N; multiset<long> S;
  REP(i, N){ cin >> A; S.insert(A); }
  long L, R, a, b; vector<vector<long>> ans;
  REP(i, N-1){ 
    L=*S.begin(); R=*S.rbegin(); a=R-L; b=L-R;
    S.erase(S.find(L)); S.erase(S.find(R));
    if(S.size()){
      long l=*S.begin(), r=*S.rbegin(), m=max(abs(a-l), abs(a-r)), n=max(abs(b-l), abs(b-r));
      if(m>=n){ S.insert(a); ans.push_back({R, L}); }else{ S.insert(b); ans.push_back({L, R}); }
    }else{ cout << a << "\n"; ans.push_back({R, L}); }
  }
  REP(i, N-1){ cout << ans[i][0] << " " << ans[i][1] << "\n"; }
}