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>
#include <stdlib.h>
#include <cmath>
using namespace std;

int main(){
    int N;
    vector<int> d;
    cin>>N;

    for(int i=0;i<N;i++){
        int x;
        cin>>x;
        d.push_back(x);
    }
    sort(d.begin(),d.end());
    int min=d.at(0), max=d.at(N-1);

    vector<int> a;

    if(N>2){
    for(int i=1;i<N-1;i++){
        int x=d.at(i);
        if(x<0){
            a.push_back(max);
            max-=x;
            
        }
        else{
            a.push_back(min);
            min-=x;
            
        }
    }
    }
    long int ans=max-min;
    cout<<ans<<endl;

    if(N>2){
        for(int i=1;i<N-1;i++){
            cout<<a.at(i-1)<<' '<<d.at(i)<<endl;
        }
    }

    cout<<max<<' '<<min<<endl;

    


}