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 <string>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <iostream>
#include<math.h>
#include<bitset>
#include<iomanip>
#include<queue>
using namespace std;

int main() {
	int N; cin >> N;
	vector<int> A(N),B,C;
	for (int i = 0; i < N; i++)cin >> A[i];
	sort(A.begin(), A.end());
	int max = A[N - 1], min = A[0];
	for(int i=1;i<N-1;i++) {
		if (A[i] >= 0) {
			B.push_back(min); C.push_back(A[i]);
			min -= A[i];
		}
		else {
			B.push_back(max); C.push_back(A[i]);
			max -= A[i];
		}
		
	}

	cout << max-min << endl;
	for (int i = 0; i < B.size(); i++)cout << B[i] << " " << C[i] << endl;
	cout << max << " " << min << endl;
}


