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=new int [N];
    for(int i=0; i<N; i++)
        cin>>A[i];
    sort(A,A+N,greater<int>());///降序;

    int res = 0;
    res += A[0];

    for (int i = 1; i < N-1; i++)
        res += abs(A[i]);

    res -= A[N-1];
    cout <<res << '\n';
    int i=1;
    for (; i < N - 1 && A[i] >= 0; i++)
    {
        cout << A[N-1] << ' ' << A[i] << '\n';
        A[N-1] -= A[i];
    }

    cout << A[0] << ' ' << A[N-1] << '\n';
    A[0] -= A[N-1];

    for (; i < N - 1; i++)
    {
        cout << A[0] << ' ' << A[i] << '\n';
        A[0] -= A[i];
    }
    return 0;
}
