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<vector>
#include<algorithm>
#include<utility>
#define lli long long
#define all(i) i.begin(),i.end()
#define rall(i) i.rbegin(),i.rend()
#define rep0(i,j) for(int i=0;i<j;i++)
#define rep1(i,j) for(int i=1;i<=j;i++)
#define rep0d(i,j) for(int i=j-1;i>=0;i--)
#define MAX 1000000007

using namespace std;

int main() {
	int n,plus=0,minus=0, zero=0;
	cin >> n;
	vector<lli> vec(n);

	rep0(i, n) {
		cin >> vec[i];
		if (vec[i] > 0)plus++;
		else if (vec[i] < 0)minus++;
		else zero++;

	}

	if (zero + plus == n) {
		sort(all(vec));
		lli ans = -vec[0];
		rep1(i, n - 1)ans += vec[i];
		cout << ans << '\n';

		rep0(i, n - 2) {
			cout << vec[i] << ' ' << vec[i + 1] << '\n';
			vec[i + 1] = vec[i] - vec[i + 1];
		}
		cout << vec[n - 1] << ' ' << vec[n - 2] << '\n';

	}
	else if (zero + minus == n) {
		sort(rall(vec));
		lli ans = vec[0];
		rep1(i, n - 1)ans -= vec[i];
		cout << ans << '\n';

		rep0(i, n - 2) {
			cout << vec[i] << ' ' << vec[i + 1] << '\n';
			vec[i + 1] = vec[i] - vec[i + 1];
		}
		cout << vec[n - 1] << ' ' << vec[n - 2] << '\n';

	}
	else if (zero > 0) {
		lli ans = 0;
		sort(rall(vec));
		rep0(i, plus)ans += vec[i];
		rep0(i, n-plus)ans -= vec[i + plus];
		cout << ans << '\n';

		rep0(i, plus) {
			cout << vec[plus-i] << ' ' << vec[plus - i - 1] << '\n';
			vec[plus - i - 1] = vec[plus - i] - vec[plus - i - 1];
		}
		cout << vec[0] << ' ' << vec[1] << '\n';
		vec[plus] = vec[0] - vec[1];

		rep0(i, n - plus - 1) {
			cout << vec[plus + i] << ' ' << vec[plus + i + 1] << '\n';
			vec[plus + i + 1] = vec[plus + i] - vec[plus + i + 1];
		}
	}
	else {
		sort(rall(vec));
		lli ans = 0;
		rep0(i, plus)ans += vec[i];
		rep0(i, n - plus)ans -= vec[i + plus];
		cout << ans << '\n';

		rep0(i, plus-1) {
			cout << vec[plus - i] << ' ' << vec[plus - i - 1] << '\n';
			vec[plus - i - 1] = vec[plus - i] - vec[plus - i - 1];
		}
		cout << vec[0] << ' ' << vec[1] << '\n';
		vec[plus-1] = vec[0] - vec[1];

		rep0(i, n - plus - 1) {
			cout << vec[plus -1 + i] << ' ' << vec[plus + i] << '\n';
			vec[plus + i] = vec[plus + i-1] - vec[plus + i];
		}

	}


}