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 <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <cstdlib>

using namespace std;

#define LL long long
#define INF 0x7fffffff
#define P 998244353

inline LL read(){
    LL x=0,f=1; char ch=getchar();
    while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    return x*f;
}

int a[100010],b[100010],cnta,cntb;
int jian[100010],jia[100010],cntjian,cntjia;
int main(){
	int n=read();
	for (int i=1; i<=n; i++){
		int x=read();
		a[i]=x;
	}
	sort(a+1,a+n+1);
	// sort(b+1,b+cntb+1);
	int njian = n/2;
	int ans=0;
	for (int i=1; i<=njian; i++) jian[++cntjian]=a[i],ans-=a[i];
	for (int i=njian+1; i<=n; i++) jia[++cntjia]=a[i],ans+=a[i];
	printf("%d\n",ans);
	int now = 0;
	while (1){
		if(cntjian && cntjia){
			int tjia = jia[cntjia]; cntjia--;
			int tjian = jian[cntjian]; cntjian--;
			int p;
			if (cntjia<cntjian) p = tjia-tjian, jia[++cntjia]=p,printf("%d %d\n",tjia,tjian);
			else p = tjian-tjia,jian[++cntjian]=p,printf("%d %d\n",tjian,tjia);
			now^=1;
			n--;
		}
		if (n==1) break; 
	}
	return 0;
}