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;
#define ss second
#define ff first
#define pb push_back
#define mp make_pair
int x[100100];
vector<pair<int,int> > v;
int main() {
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
        cin>>x[i];
    sort(x,x+n);
    ll b=x[0];
    for(int i=1;i<n-1;i++)
    {   if(x[i]>=0){
     v.pb(mp(b,x[i]));
     b-=x[i];}
     else
     {
         v.pb(mp(x[n-1],x[i]));
         x[n-1]-=x[i];
     }
    }
    v.pb(mp(x[n-1],b));
    x[n-1]-=b;
    cout<<x[n-1]<<endl;
    for(int i=0;i<n-1;i++)
        cout<<v[i].ff<<' '<<v[i].ss<<endl;
    return 0;
}
