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(void){
    // Your code here!
    int n,ans=0,p;
    cin >> n;
    vector<int> a(n);
    for (int i=0;i<n;i++){
        cin >> a.at(i);
        if (a.at(i)<=0) p=a.at(i)*(-1);
        else p=a.at(i);
        ans+=p;
    }
    sort(a.begin(),a.end());
    if (a.at(0)<=0 && a.at(n-1)>=0) cout << ans << endl;
    else if (a.at(n-1)<0) cout << ans+a.at(n-1)*2 << endl;
    else cout << ans-a.at(0)*2 << endl;
    int t=a.at(0);
    for (int i=1;i<n-1;i++){
        if (a.at(i)>=0) {cout << t << " " << a.at(i) << endl;
        t=t-a.at(i);}
    }
    cout << a.at(n-1) << " " << t << endl;
    t=a.at(n-1)-t;
    for (int i=1;i<n-1;i++){
        if (a.at(i)<0) cout << t << " " << a.at(i) << endl;
        t=t-a.at(i);
    }
}