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: Wrong Answer)

#include<bits/stdc++.h>
#include<iostream>
#include<fstream>
#define ll long long
using namespace std;

int main(){
    int N;  cin >> N;
    vector<int> A(N);
    for(int i=0; i<N; i++) cin >> A[i];
    sort(A.begin(), A.end());

    vector<int> ans(N,0);
    int S = 0;
    for(int i=0; i<N; i++){
        if((N%2==0 && i%2==0) || (N%2==1 && i%2==1)){
            ans[i] = A[i/2];
            S -= ans[i];
        }else{
            ans[i] = A[i/2+N/2];
            S += ans[i];
        }
    }


    cout << S << endl;
    int y = ans[0];
    for(int i=1; i<N; i++){
        cout << ans[i] << " " << y << endl;
        y = ans[i] - y;
    }
}