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 <algorithm>
#include <cmath>
#include <functional>
#include <iostream>
#include <map>
#include <numeric>
#include <stack>
#include <string>
#include <vector>
typedef long long ll;
using namespace std;
const int MOD = (int)1e9 + 7;
const ll INF = (ll)1e18 + 7;
#define int long long
#define vint vector<int>
#define pint pair<int, int>
#define pb(a) push_back(a)
#define all(v) v.begin(), v.end()
#define dbg(x) cerr << #x << " = " << (x) << endl
#define yn(b) cout << ((b) ? "Yes" : "No") << endl
#define YN(b) cout << ((b) ? "YES" : "NO") << endl

int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int N;
	cin >> N;

	vint A(N);
	vector<pint> ans, ans2;

	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}

	sort(all(A));
	int mae;

	for (int i = 0; i < N - 1; i++) {
		int k1, k2;

		//(小さいほう)-(大きいほう)
		if (i == 0) {
			k1 = A[0];
			k2 = A[1];
			mae = k1 - k2;
		} else if (i == N - 2) {
			k2 = min(A[i + 1], mae);
			k1 = max(A[i + 1], mae);
			mae = k1 - k2;
		} else {
			k1 = min(A[i + 1], mae);
			k2 = max(A[i + 1], mae);
			mae = k1 - k2;
		}
		pint tmp;
		tmp.first = k1;
		tmp.second = k2;
		ans.push_back(tmp);
	}

	//take2
	int mae2;
	sort(A.rbegin(), A.rend());
	for (int i = 0; i < N - 1; i++) {
		int k1, k2;

		//(大)-(小)
		if (i == 0) {
			k1 = A[0];
			k2 = A[1];
			mae2 = k1 - k2;
		} else if (i == N - 2) {
			k2 = min(A[i + 1], mae2);
			k1 = max(A[i + 1], mae2);
			mae2 = k1 - k2;
		} else {
			k1 = max(A[i + 1], mae2);
			k2 = min(A[i + 1], mae2);
			mae2 = k1 - k2;
		}
		pint tmp;
		tmp.first = k1;
		tmp.second = k2;
		ans2.push_back(tmp);
	}
	if (mae >= mae2) {
		cout << mae << endl;
		for (int i = 0; i < N - 1; i++) {
			cout << ans[i].first << " " << ans[i].second << endl;
		}
	} else {
		cout << mae2 << endl;
		for (int i = 0; i < N - 1; i++) {
			cout << ans2[i].first << " " << ans2[i].second << endl;
		}
	}

	return 0;
}