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
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
using namespace std;
typedef pair<ll,ll> P;
const ll MAX_N=1e5;
ll A[MAX_N],N;
			
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>N;
	REP(i,N)cin>>A[i];
	sort(A,A+N);
	ll ans=0;
	ans-=A[0];
	ans+=A[N-1];
	REP(i,N-2){
		ll a=A[i+1];
		ans+=llabs(a);
	}
	cout<<ans<<endl;
	ll ite=1;
	FOR(i,1,N-1){
		if(A[i]<0){
			ite++;
		}else{
			break;
		}
	}
	FOR(i,ite,N-1){
		cout<<A[0]<<' '<<A[i]<<endl;
		A[0]-=A[i];
	}
	REP(i,ite){
		cout<<A[N-1]<<' '<<A[i]<<endl;
		A[N-1]-=A[i];
	}
}