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>
#define  MAX 100005
#define  MOD 1000000007
#define ll long long
using namespace std;

ll arr[100005];

int main() {
    int n;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>arr[i];
    }
    sort(arr+1,arr+1+n);
    ll ans=arr[n]-arr[1];
    for(int i=2;i<n;i++){
        if(arr[i]>0)ans+=arr[i];
        else ans-=arr[i];
    }
    cout<<ans<<endl;
    for(int i=2;i<n;i++){
        if(arr[i]>0){
            cout<<arr[1]<<" "<<arr[i]<<endl;
            arr[1]-=arr[i];
        }else{
            cout<<arr[n]<<" "<<arr[i]<<endl;
            arr[n]-=arr[i];
        }
    }
    cout<<arr[n]<<" "<<arr[1]<<endl;arr[n]-=arr[1];
}
