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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

static const int INF=1e9;
static const int MOD=1000000007;
typedef long long ll;
typedef pair<int,int> P;

int main(){
    int N; cin >> N;
    int a[N];
    for(int i=0;i<N;i++) cin >> a[i];
    sort(a,a+N);
    vector<pair<int,int>> ans;
    int i=N-1;
    while(true){
        if(a[i]<0||i==1) break;
        ans.push_back({a[0],a[i]});
        a[0]=a[0]-a[i];
        i--;
    }
    int j=0;
    while(true){
        if((i-j)==1) break;
        ans.push_back({a[i],a[j]});
        a[i]=a[i]-a[j];
        j++;
    }
    ans.push_back({a[j],a[j+1]});
    cout << a[j+1]-a[j] << endl;
    for(int k=0;k<N-1;k++){
        cout << ans[k].first << " " << ans[k].second << endl;
    }
}