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 rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main(){
    int n;
    cin>>n;
    vector<int>a(n);
    rep(i,n)cin>>a[i];
    sort(a.begin(),a.end());
    vector<int>p,m;
    p.push_back(a[n-1]);
    m.push_back(a[0]);
    for(int i=1; i<n-1; ++i){
        if(a[i]<0)m.push_back(a[i]);
        else p.push_back(a[i]);
    }
    vector<P>ans;
    for(int i=1; i<p.size(); ++i){
        ans.push_back(P(m[0],p[i]));
        m[0]-=p[i];
    }
    for(int i=0; i<m.size(); ++i){
        ans.push_back(P(p[0],m[i]));
        p[0]-=m[i];
    }
    cout<<p[0]<<endl;
    rep(i,ans.size()){
        cout<<ans[i].first<<" "<<ans[i].second<<endl;
    }
}