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>
#define int long long
using namespace std;
vector<int> a;
int n,x;
signed main()
{
	cin >> n;
	for(int i = 1; i <= n;i++)
	{
		int x;
		cin >> x;
		a.push_back(x);
	}
	sort(a.begin(), a.end());
	int mn, mx;
	mn = a[0]; mx = a[n-1];
	for(int i = 1; i < n - 1;i++)
	{
		if(a[i] < 0) mx = mx - a[i];
		else mn = mn - a[i];
	}
	cout << mx - mn << "\n";
	mn = a[0]; mx = a[n-1];
	for(int i = 1; i < n - 1;i++)
	{
		if(a[i] < 0){ 
		    cout << mx << " " << a[i] << "\n";
		    mx -= a[i];
		}
		else{ 
		    cout << mn << " " << a[i] << "\n";
		    mn -= a[i];
		}
	}
	cout << mx << " " << mn;
}