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 FOR(i, a, b) for (int i=(a); i<(b); i++)
#define PPC(x) __builtin_popcount((x))
#define ALL(x) (x).begin(), (x).end()
#define pb push_back
using namespace std;

const int maxN = 100042;

int T[maxN];
vector <pair <int, int> > res;

int main()
{
	int n;
	scanf ("%d", &n);
	
	FOR(i, 0, n)
		scanf ("%d", T+i);
	sort(T, T+n);
		
	int i;
	for (i=1; T[i]<0; i++)
		res.pb({T[n-1], T[i]}), T[n-1] -= T[i];
	for (; i<n-1; i++)
		res.pb({T[0], T[i]}), T[0] -= T[i];
	res.pb({T[n-1], T[0]});
	T[n-1] -= T[0];
	
	printf("%d\n", T[n-1]);
	for (auto p : res)
		printf("%d %d\n", p.first, p.second);
	
	return 0;
}
