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 <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <bitset>
#include<vector>
using namespace std;
typedef long long ll;
#define rep(i,k) for(i=0;i<k;i++)


int main() {
	ll a[100000], b = 0 , i, j, k, l, n, z = 100, sum = 0;
	cin >> n;
	rep(i, n) {
		cin >> a[i];
		b = b + abs(a[i]);
	}
	sort(a, a + n);
	if (a[0] >= 0)cout << b - 2*a[0] << "\n";
	else {
		if (a[n - 1] < 0)cout << b + 2 * a[n - 1] << "\n";
		else cout << b << "\n";
	}

	k = a[0]; l = a[n - 1];
	for (i = 1; i < n - 1; i++) {
		if (a[i] < 0) {
			cout << l << " " << a[i] << "\n";
			l = l - a[i];
		}
		else {
			cout << k << " " << a[i] << "\n";
			k = k - a[i];
		}
	}
		cout << l << " " << k << "\n";

	return 0;
}