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;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++)cin>>a[i];
    sort(a,a+n);
    vector<pair<int,int> > ans;
    int ma=a[n-1],mi=a[0];
    for(int i=1;i<n-1;i++){
        if(a[i]<0){
            ans.push_back(make_pair(ma,a[i]));
            ma-=a[i];
        }
        else{
            ans.push_back(make_pair(mi,a[i]));
            mi-=a[i];
        }
    }
    ans.push_back(make_pair(ma,mi));
    cout << ma-mi <<endl;
    for(int i=0;i<n-1;i++){
        cout <<ans[i].first<<" "<<ans[i].second<<endl;
    }
    return 0;
}