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;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define P pair<ll,ll>

int main(){
  int n;
  cin>>n;
  vector<int> a(n);
  rep(i,n) cin>>a[i];
  sort(a.begin(),a.end());
  int mi=a[0], ma=a[n-1];
  vector<int> plus,minus;
  rep(i,n){
    if(i==0||i==n-1) continue;
    if(a[i]>0) plus.push_back(a[i]);
    else minus.push_back(a[i]);
  }
  vector<P> ans;
  int m=mi;
  for(auto x : plus){
    ans.push_back(make_pair(m,x));
    m=m-x;
  }
  ans.push_back(make_pair(ma,m));
  m=ma-m;
  for(auto x : minus){
    ans.push_back(make_pair(m,x));
    m=m-x;
  }
  cout<<m<<endl;
  for(auto p : ans){
    cout<<p.first<<" "<<p.second<<endl;
  }
}