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 long long ll;
const int MOD=1000000007;
#define INF 1LL<<30
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()

int main(){
    int n;
    cin>>n;

    vector<int> a(n);
    rep(i,n) cin>>a[i];
    sort(all(a));

    int sum=0;
    rep(i,n) sum+=abs(a[i]);
    if(a[0]>0) sum-=2*a[0];
    if(a[n-1]<0) sum+=2*a[n-1];
    cout<<sum<<endl;

    int acc=a[0];

    for(int i=1;i<n-1;i++){
        if(a[i]>=0){
            cout<<acc<<" "<<a[i]<<endl;
            acc-=a[i];
        }
    }

    cout<<a[n-1]<<" "<<acc<<endl;
    acc=a[n-1]-acc;

    for(int i=1;i<n-1;i++){
        if(a[i]<0){
            cout<<acc<<" "<<a[i]<<endl;
            acc-=a[i];
        }
    }


}