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 i, n, res1 = 0, res2 = 0;
    cin >> n;
    int a[n];
    for(i=0; i<n; i++){
        cin >> a[i];
    }
    sort(a, a+n);
    res2 = a[n-1];
    res1 = a[0];
    for(i=1; i<n-1; i++){
        if(a[i] < 0){
            res2 -= a[i];
        }else{
            res1 -= a[i];
        }
    }
    cout << res2 - res1 << '\n';
    res2 = a[n-1];
    res1 = a[0];
    for(i=1; i<n-1; i++){
        if(a[i] < 0){
            cout << res2 << ' ' << a[i] << '\n';
            res2 -= a[i];
        }else{
            cout << res1 << ' ' << a[i] << '\n';
            res1 -= a[i];
        }
    }
    cout << res2 << ' ' << res1;
    return 0;
}