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;


int main()
{
    int n;
    vector <int> vi, vj;

    cin>>n;
    int a[n+5];

    int p=0;
    for(int i=0; i<n; i++)
    {
        cin>>a[i];
        if(a[i]<0) p++;
    }
    if(!p) p++;
    if(p==n) p--;

    sort(a,a+n);

    for(int i=p; i<n-1; i++)
    {
        vi.push_back(a[0]);
        vj.push_back(a[i]);
        a[0]-=a[i];
    }

    for(int i=0; i<p; i++)
    {
        vi.push_back(a[n-1]);
        vj.push_back(a[i]);
        a[n-1]-=a[i];
    }

    cout<<a[n-1]<<endl;
    for(int i=0; i<vi.size(); i++)
        cout<<vi[i]<<" "<<vj[i]<<endl;

    return 0;
}
