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 <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <stdio.h>
#include <cmath>

using namespace std;

typedef long long ll;
int a[100010];
int ans[100010];
int main() {
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	sort(a + 1, a + n + 1);
	int maxn = a[n];
	int minn = a[1];
	for (int i = 2; i <= n-1; i++) {
		if (a[i] > 0) {
			ans[i] = minn;
			minn -= a[i];
		}
		else {
			ans[i] = maxn;
			maxn -= a[i];
		}
	}
	cout << maxn - minn << endl;
	for (int i = 2; i <= n - 1; i++) {
		cout << ans[i] << " " << a[i] << endl;
	}
	cout << maxn << " " << minn << endl;

	return 0;
}