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;

typedef long long ll;
const ll InF = 1e15;

int main() {
 int n;
 cin>>n;
 vector<ll> a(n);
 for (int i = 0; i < n; i++) cin>>a[i];
 sort(a.begin(),a.end());
 vector<pair<ll,ll>> ps;
 int k = 0;
 for (int i = 0; i < n - 1; i++) {
   if (a[i] < 0 && a[i + 1] < 0) {
     ps.emplace_back(a[n - 1],a[i]);
     k++;
     a[n - 1] = a[n - 1] - a[i];
   } else break;
 }
 ll res = a[k];
 for (int i = k + 1; i < n; i++) {
   if (i == n - 1) {
     ps.emplace_back(a[i],res);
     res = a[i] - res;
   } else {
     ps.emplace_back(res,a[i]);
     res -= a[i];
   }                            
 }
 cout<<res<<endl;
 for (int i = 0; i < n - 1; i++) {
   cout<<ps[i].first<<" "<<ps[i].second<<endl;;
 }
 return 0;
}