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<bits/stdc++.h>
using namespace std;
inline int_fast64_t read(){int_fast64_t x=0,f=1;char c=getchar();while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return x*f;}
inline string read_s(){string s="";char c=getchar();while(c<'!'||c>'~'){c=getchar();}while(c>='!'&&c<='~'){s+=c;c=getchar();}return s;}
void Main(){
    int N = read();
    vector<int> A(N);
    for (auto& a:A) {
        a = read();
    }
    sort(A.begin(), A.end());
    vector<pair<int, int>> Ans(N-1);
    for (int i = 1; i<N-1; i++) {
        if (A[i] > 0) {
            Ans[i-1] = make_pair(A[0], A[i]);
            A[0] -= A[i];
        } else {
            Ans[i-1] = make_pair(A[N-1], A[i]);
            A[N-1] -= A[i];
        }
    }
    Ans[N-2] = make_pair(A[N-1], A[0]);
    printf("%d\n", A[N-1]-A[0]);
    for (auto ans:Ans) {
        printf("%d %d\n", ans.first, ans.second);
    }
}
int main(){cin.tie(nullptr);ios::sync_with_stdio(false);Main();return 0;}