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 <iostream>
#include <algorithm>
#include <vector>
#include <string>
#define int long long
using namespace std;
signed main() {
	int m; cin >> m;
	int a[123456], p = m - 2;
	for (int i = 0; i < m; i++)cin >> a[i]; sort(a, a + m, greater<int>());
	for (int i = 0; i < m - 1; i++) {
		if (a[i + 1] <= 0) {
			p = i;
			break;
		}
	}
	int ans = a[p + 1];
	for (int i = p; i > 0; i--) {
		ans -= a[i];
	}
	ans = a[0] - ans;
	for (int i = p + 2; i < m; i++)ans -= a[i];
	cout << ans << endl;
	ans = a[p + 1];
	for (int i = p; i > 0; i--) {
		cout << ans << ' ' << a[i] << endl;
		ans -= a[i];
	}
	cout << a[0] << ' ' << ans << endl;
	ans = a[0] - ans;
	for (int i = p + 2; i < m; i++) {
		cout << ans << ' ' << a[i] << endl;
		ans -= a[i];
	}
	cin >> m;
}