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;
typedef pair<int,int> pii;
#define pb push_back
#define mp make_pair
#define rep(i,n) for(int i=0;i<(n);++i)
constexpr int mod=1000000007;
constexpr int mod1=998244353;
vector<int> dx={0,-1,0,1},dy={1,0,-1,0};



int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;cin >> n;
	vector<int> a(n);
	rep(i,n) cin >> a.at(i);
	sort(a.begin(),a.end());
	vector<pii> ans;
	for(int i=1;i<n-1;++i){
		if(a.at(i)>=0){
			ans.pb(mp(a.at(0),a.at(i)));
			a.at(0)-=a.at(i);
		}
		else{
			ans.pb(mp(a.at(n-1),a.at(i)));
			a.at(n-1)-=a.at(i);
		}
	}
	ans.pb(mp(a.at(n-1),a.at(0)));
	a.at(n-1)-=a.at(0);
	cout << a.at(n-1) << endl;
	rep(i,n-1) cout << ans.at(i).first << " " << ans.at(i).second << endl;
}
