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;

typedef pair<int, int> P;
typedef long long int ll;

int N, A[100000];

int main(){
    cin >> N;
    for(int i=0;i<N;i++)    cin >> A[i];

    sort(A, A+N);
    vector<P> ans;
    int M;
    for(int i=1;i<N-1;i++){
        if(A[i] < 0){
            ans.push_back( P(A[N-1], A[i]) );
            A[N-1] = A[N-1] - A[i];
        } else {
            ans.push_back( P(A[0], A[i]) );
            A[0] = A[0] - A[i];
        }
    }

    M = A[N-1] - A[0];

    cout << M << endl;
    for(auto i : ans)   cout << i.first << " " << i.second << endl;
    cout << A[N-1] << " " << A[0] << endl;
}