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)

#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include<iostream>
#include<queue>
#include<vector>

using namespace std;
using i32 = int_fast32_t;
using i64 = int_fast64_t;
#define rep(i, n) for (i32 i = 0; i < (i32)(n); i++)
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
using P = pair<i64,i64>;

int main(){
ios::sync_with_stdio(false);
std::cin.tie(nullptr);
i64 n;
cin >> n;
priority_queue<i64,vector<i64>, greater<i64>> que;
rep(i,n){
    i64 a;
    cin >> a;
    que.push(a);
}
vector<P> rec;
while(que.size() > 1){
    i64 a = que.top();
    que.pop();
    i64 b = que.top();
    que.pop();
    if(que.size() >= 1) {
        if (a > 0) {
            que.push(b - a);
            rec.emplace_back(P(b, a));
        } else if (a < 0) {
            que.push(a - b);
            rec.emplace_back(P(a, b));
        } else if (a >= 0) {
            que.push(b - a);
            rec.emplace_back(P(b, a));
        }
    }else{
        que.push(b - a);
        rec.emplace_back(P(b,a));
    }
}
cout << que.top() << endl;
for(auto v: rec){
    cout << v.first << " " << v.second << endl;
}
}