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<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<map>
#include<iomanip>
#include<set>
#define rep(index,num) for(int index=0;index<num;index++)
#define rep1(index,num) for(int index=1;index<=num;index++)
#define scan(argument) cin>>argument
#define prin(argument) cout<<argument<<endl
#define kaigyo cout<<endl
#define eps 1e-7
#define mp(a1,a2) make_pair(a1,a2)
typedef long long ll;
typedef long double ld;
using namespace std;
typedef pair<ll,ll> pll;
typedef pair<int,int> pint;
typedef vector<int> vint;
typedef vector<ll> vll;
typedef vector<pint> vpint;
typedef vector<pll> vpll;
ll INFl=1e+18+1;
int INF=1e+9+1;
int main(){
	int N,A[100001];
	scan(N);
	rep(i,N){
		scan(A[i]);
	}
	sort(A,A+N);
	int M=0;
	vint p,m;
	rep(i,N){
		if(i==0) M-=A[i];
		else if(i==N-1) M+=A[i];
		else{
			M+=abs(A[i]);
			if(A[i]>=0) p.push_back(A[i]);
			else m.push_back(A[i]);
		}
	}
	prin(M);
	int sum1=A[0];
	rep(i,p.size()){
		printf("%d %d\n",sum1,p[i]);
		sum1-=p[i];
	}
	int sum2=A[N-1];
	rep(i,m.size()){
		printf("%d %d\n",sum2,m[i]);
		sum2-=m[i];
	}
	printf("%d %d\n",sum2,sum1);
	return 0;
}