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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<stdio.h>
#include<string>
#include<vector>
#include<map>
#include<tuple>
#include<algorithm>
#include<cmath>
#include<limits>
#include<set>
#include<deque>
#include<queue>
#include<stack>
using namespace std;
#define int long long int
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define dup(x,y) (((x)+(y)-1)/(y))
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
typedef pair<int, int>P;
const int MOD = 1e9 + 7;
//const int MOD = 998244353;
const int INF = 1e18;
const long double PI = (acos(-1));


signed main() {
	int N;
	cin >> N;
	vector<int>A(N);
	rep(i, N)cin >> A[i];
	sort(A.begin(), A.end());
	int M = 0;
	rep(i, N) {
		if (!i)M -= A[i];
		else M += A[i];
	}
	cout << M << endl;
	int x, y;
	x = A[0];
	rep(i, N - 1) {
		y = A[i + 1];
		if (i == N - 2)swap(x, y);
		cout << x << " " << y << endl;
		int res = x - y;
		x = res;
	}
}