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>
using namespace std;
#define ll long long

int main() {
	ll N, all = 0;
	cin >> N;
	vector<ll> X(N);
	for (int i = 0;i < N;i++) {
		cin >> X[i];
		all += X[i];
	}
	sort(X.begin(), X.end());
	ll ans = 0;
	for (int i = 1;i < N - 1;i++)
		ans += abs(X[i]);
	ans -= X[0];
	ans += X[N - 1];

	cout << ans << endl;
	for (int i = 1;i < N - 1;i++) {
		if (X[i] < 0) {
			cout << X[N - 1] << " " << X[i] << endl;
			X[N - 1] -= X[i];
		}
		else {
			cout << X[0] << " " << X[i] << endl;
			X[0] -= X[i];
		}
	}
	cout << X[N - 1] << " " << X[0] << endl;
}