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;
typedef long long ll;
#define int ll
inline int read()
{
	int out=0,fh=1;
	char jp=getchar();
	while ((jp>'9'||jp<'0')&&jp!='-')
		jp=getchar();
	if (jp=='-')
		fh=-1,jp=getchar();
	while (jp>='0'&&jp<='9')
		out=out*10+jp-'0',jp=getchar();
	return out*fh;
}
const int MAXN=2e5+10;
int n;
multiset<int> s;
int a[MAXN],b[MAXN],cnt=0;
signed main()
{
	n=read();
	for(int i=1;i<=n;++i)
		s.insert(read());
	int p=n;
	while(1)
	{
		if(s.size()==1)
		{
			cout<<*s.begin()<<endl;
			break;
		}
		multiset<int>::iterator it1;
		multiset<int>::iterator it2;
		it1=s.begin();
		it2=s.end();
		--it2;
		int x;
		if(p&1)
		{
			++cnt;
			a[cnt]=*it1;
			b[cnt]=*it2;
			x=*it1-*it2;
		}
		else
		{
			++cnt;
			a[cnt]=*it2;
			b[cnt]=*it1;
			x=*it2-*it1;
		}
		s.erase(it1);
		s.erase(it2);
		s.insert(x);
		--p;
	}
	for(int i=1;i<=cnt;++i)
		printf("%lld %lld\n",a[i],b[i]);
	return 0;
}