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
#include <bits/stdc++.h>
using namespace std;
int main() {
	long long int n;
	cin >> n;
	long long int a[n], ans[2][n];
	for (int i = 0; i < n; i++)
		cin >> a[i];
	sort(a, a + n);
	for (int i = 1; i <= n - 2; i++) {
		if (a[i] <= 0) {
			ans[0][i] = a[n - 1];
			ans[1][i] = a[i];
			a[n - 1] -= a[i];
		}
		else {
			ans[0][i] = a[0];
			ans[1][i] = a[i];
			a[0] -= a[i];
		}
	}
	ans[0][n - 1] = a[n - 1];
	ans[1][n - 1] = a[0];
	cout << a[n - 1] - a[0] << '\n';
	for (int i = 1; i <= n - 1; i++)
		cout << ans[0][i] << ' ' << ans[1][i] << '\n';
	return 0;
}