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}

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}

Examples

Input

3
1 -1 2


Output

4
-1 1
2 -2


Input

3
1 1 1


Output

1
1 1
1 0
#define _GLIBCXX_DEBUG
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <utility>
#include <string>
using namespace std;
typedef long long ll;
#define ALL(v) (v).begin(), (v).end()

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

		sort(A, A+N);

		vector<pair<int, int>> v;
		for (int i = 1; i < N-1; i++) {
				if (A[i] >= 0) {
						v.push_back(make_pair(A[0], A[i]));
						A[0] -= A[i];
				} else {
						v.push_back(make_pair(A[N-1], A[i]));
						A[N-1] -= A[i];
				}
		}

		v.push_back(make_pair(A[N-1], A[0]));
		A[N-1] -= A[0];

		cout << A[N-1] << endl;
		for (int i = 0; i < N-1; i++) {
				cout << v[i].first << " " << v[i].second << endl;
		}
		return 0;
}
