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
/*input
2
1 -1
*/
#include "bits/stdc++.h"
using namespace std;
long long a[100005],n;
vector<pair<int,int>> moves;

void put(long long &x, long long &y){
	moves.emplace_back(x,y);
	x-=y;
}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i) cin>>a[i];
	
	sort(a+1,a+n+1);
	if(a[1] >= 0){
		for(int i=2;i<n;++i)
			put(a[1],a[i]);
		put(a[n],a[1]);
	} else if(a[n] <= 0){
		for(int i=n-1;i>0;--i) put(a[n],a[i]);
	} else{
		int last = 1;
		while(last<n && a[last+1] < 0) ++last;

		for(int i=last;i>1;--i)
			put(a[n],a[i]);

		for(int i=last+1;i<n;++i)
			put(a[1],a[i]);
		
		put(a[n],a[1]);
	}
	cout<<moves.back().first-moves.back().second<<'\n';
	for(auto i:moves)
		cout<<i.first<<' '<<i.second<<'\n';
}