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>
using namespace std;
const int MAXN=100000+10;
int n;
int a[MAXN];
int m;
int maxn,t;
int res;
struct answer{
	int x,y;
}ans[MAXN];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
		if(a[i]>maxn){
			maxn=a[i];
			t=i;
		}
	}
	for(int i=t;i<=n;i++)
		a[i]=a[i+1];
	sort(a+1,a+n);
	res=a[1];
	for(int i=2;i<=n-1;i++){
		ans[i-1].x=res;ans[i-1].y=a[i];
		res-=a[i];
	}
	ans[n-1].x=maxn;ans[n-1].y=res;
	res=max(maxn-res,res-maxn);
	printf("%d\n",res);
	for(int i=1;i<=n-1;i++)
		printf("%d %d\n",ans[i].x,ans[i].y);
	return 0;
}