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 <vector>
#include <algorithm>

using namespace std;

int main(){
    int N;
    while(cin >> N){
        vector<int> a(N);
        for(auto& t : a) cin >> t;
        sort(a.begin(), a.end());
        vector<pair<int,int>> res;
        for(int i=1;i<a.size()-1;i++){
            if(a[i] < 0){
                res.emplace_back(a.back(), a[i]);
                a.back() -= a[i]; 
            } else {
                res.emplace_back(a[0], a[i]);
                a[0] -= a[i];
            }
        }
        res.emplace_back(a.back(), a[0]);
        cout << a.back() - a[0] << endl;
        for(auto& p : res) cout << p.first << " " << p.second << endl;
    }
}