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<stdio.h>
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define f_(i,a,b) for (int i=a;i>=b;i--)
#define f(i,a,b) for (int i=a;i<=b;i++)
const ll mod=1000000007;

int n,a[100005];
vector<pair<int,int> > ans;
int main() {
//	freopen("","r",stdin);
//	freopen("","w",stdout);
	scanf("%d", &n);
	f(i,1,n) scanf("%d", &a[i]);
	sort(a+1,a+1+n);
	f_(i,n-1,2) {
		if (a[i]<0) break;
		ans.push_back(make_pair(a[1],a[i]));
		a[1]-=a[i];
	}
	f(i,2,n-1) {
		if (a[i]>=0) break;
		ans.push_back(make_pair(a[n],a[i]));
		a[n]-=a[i];
	}
	ans.push_back(make_pair(a[n],a[1]));
	a[n]-=a[1];
	printf("%d\n", a[n]);
	for (int i=0;i<ans.size();i++) printf("%d %d\n", ans[i].first, ans[i].second);
	return 0;
}