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 <iostream>
#include <algorithm>
#include <cmath>

using namespace std;

const int maxn = 1e5+5 ;
int arr[maxn] ;
int main()
{
    int n ;
    cin >> n ;
    int l = 0 ;
    for (int i = 0 ; i < n ; i++) {
        cin >> arr[i] ;
        if (arr[i]<0)  l++ ;
    }
    sort(arr,arr+n) ;
    int ans = 0 ;
    for(int i = 1 ; i < n-1 ; i++) ans += abs(arr[i]) ;
    ans-=arr[0] ;
    ans+=arr[n-1] ;
    cout << ans << endl ;
    if (l==0) l = 1 ;
    else if (l==n) l=n-1 ;
    for(int i = l ; i < n-1 ; i++) {
        cout << arr[0] << " "<< arr[i] <<endl;
        arr[0] -=arr[i] ;
    }
    for(int i = 0 ; i < l ; i++) {
        cout << arr[n-1] << " " <<arr[i] << endl ;
        arr[n-1] -=arr[i] ;
    }
    return 0;
}
