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;
typedef long long ll;


int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr), cout.tie(nullptr);

	int n;
	cin >> n;
	vector<ll> a(n);
	for(ll &x: a) cin >> x;
	sort(a.begin(), a.end());



	{
		ll ans = -a[0] + a.back();
		for(int i = 1; i + 1 < n; ++i)
			ans += abs(a[i]);
		cout << ans << endl;
	}
/*

	int cntNeg = 1;
	for(int i = 1; i + 1 < n; ++i)
		cntNeg += a[i] < 0;*/

	ll s1 = a.back();
	for(int i = 1; i + 1 < n && a[i] < 0; ++i) {
		cout << s1 << ' ' <<a[i] << '\n';
	s1 -= a[i];
	}

	ll s2 = a[0];
	for(int i = n - 2; i > 0 && a[i] >= 0; --i) {
		cout << s2 << ' '<<a[i] << '\n';
		s2 -= a[i];
	}
	cout << s1 << ' ' << s2 << endl;
	return 0;
}
