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 <stdio.h>
#include <algorithm>
using namespace std;
int a[100010],flag[100010];

int main()
{
    int n;scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]);
    sort(a+1,a+n+1);

    flag[1]=-1;flag[n]=1;
    for (int i=2;i<n;i++)
    if (a[i]<0) flag[i]=-1;else flag[i]=1;

    int ans=0;
    for (int i=1;i<=n;i++) ans+=flag[i]*a[i];
    printf("%d\n",ans);
    for (int i=2;i<n;i++)
    {
        if (flag[i]==1)
        {
            printf("%d %d\n",a[1],a[i]);
            a[1]-=a[i];
        }
    }
    for (int i=1;i<n;i++)
    {
        if (flag[i]==-1)
        {
            printf("%d %d\n",a[n],a[i]);
            a[n]-=a[i];
        }
    }
}
