Problem:
Score : 500 points

Problem Statement

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}

Sample Input 1

3
1 -1 2

Sample Output 1

4
-1 1
2 -2

If we choose x = -1 and y = 1 in the first operation, the set of integers written on the blackboard becomes (-2, 2).

Then, if we choose x = 2 and y = -2 in the second operation, the set of integers written on the blackboard becomes (4).

In this case, we have 4 as the final integer. We cannot end with a greater integer, so the answer is 4.

Sample Input 2

3
1 1 1

Sample Output 2

1
1 1
1 0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <string>
#include <vector>
#include <list>
#include <iomanip>
#include <queue>
#include <deque>
#include <map>
#include <unordered_map>
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define repr(e,x) for(auto& e:x)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int N;
int A[100000];

int main(){
    cin>>N;
    rep(i,N) cin>>A[i];

    int plus=0, minus=0;
    rep(i,N){
        if(A[i]>=0) plus++;
        else minus++;
    }
    if(plus==0 || minus==0){
        if(plus==0){
            plus++;
            minus--;
        }else{
            plus--;
            minus++;
        }
    }
    
    sort(A,A+N);

    vector<P> x;
    int ans=A[minus-1];
    minus--;
    while(plus>1){
        x.push_back(P(ans,A[minus+plus]));
        ans-=A[minus+plus];
        plus--;
    }
    x.push_back(P(A[minus+plus],ans));
    ans=A[minus+plus]-ans;
    while(minus>0){
        x.push_back(P(ans,A[minus-1]));
        ans-=A[minus-1];
        minus--;
    }

    cout<<ans<<endl;
    repr(e,x){
        cout<<e.first<<' '<<e.second<<endl;
    }
}