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>
#define rep(i,l,r) for (int i=l;i<=r;i++)
using namespace std;
typedef long long ll;
int a,b,c,ans,N,K,M,tt,t,q[200007],ansx[200007],ansy[200007];
int main(){
	scanf("%d",&N);
	rep(i,1,N) scanf("%d",&q[i]);
	sort(q+1,q+N+1);
	int p=N; while (p>2&&q[p-1]>0) p--;
	int s=q[1];
	rep(i,p,N-1){
		ansx[++t]=s;ansy[t]=q[i];
		s-=q[i];
	}
	ansx[++t]=q[N];ansy[t]=s;
	s=q[N]-s;
	rep(i,2,p-1){
		ansx[++t]=s;ansy[t]=q[i];
		s-=q[i];
	}
	printf("%d\n",s);
	rep(i,1,t) printf("%d %d\n",ansx[i],ansy[i]);
	return 0;
}
