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;
typedef pair<int,int> P;
int main(){
    int n;
    cin>>n;
    vector<int> a(n);
    for(int i=0;i<n;i++)cin>>a.at(i);
    sort(a.begin(),a.end());
    queue<P> Q;
        for(int i=0;i<n-2;i++){
        if(a.at(i)<=0&&a.at(i+1)<=0){
            Q.push(P(a.at(n-1),a.at(i)));
           a.at(n-1)-=a.at(i);
        }
        else if(a.at(i+1)>0){
            Q.push(P(a.at(i),a.at(i+1)));
            a.at(i+1)=a.at(i)-a.at(i+1);
        }
    }
    Q.push(P(a.at(n-1),a.at(n-2)));
    cout<<a.at(n-1)-a.at(n-2)<<endl;
    while(!Q.empty()){
        P q=Q.front();
        Q.pop();
       cout<<q.first<<" "<<q.second<<endl;
    }
}