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<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<unordered_set>
#include<unordered_map>
using namespace std;
typedef long long ll;
#define mod 1000000007
#define mad(a,b) a=(a+b)%mod;
#define N 100010
typedef pair<ll,ll> P;
vector<P> v;
ll n,a[N];
int main(){
  cin>>n;
  for(int i=0;i<n;i++)cin>>a[i];
  sort(a,a+n);
  for(int i=1;i<n-1;i++){
      if(a[i]<0)continue;
      v.push_back(make_pair(a[0],a[i]));
      a[0]=a[0]-a[i];
  }
  for(int i=1;i<n-1;i++){
      if(a[i]>=0)continue;
      v.push_back(make_pair(a[n-1],a[i]));
      a[n-1]=a[n-1]-a[i];
  }
  v.push_back(make_pair(a[n-1],a[0]));
  cout<<a[n-1]-a[0]<<endl;
  for(int i=0;i<n-1;i++){
      cout<<v[i].first<<" "<<v[i].second<<endl;
  }
  return 0;
}
