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 int long long
typedef long long ll;
#define sqr(x) ((x)*(x))
#define pb push_back 
const int N=1000002,M=998244353;
int pw(int x,int y){int z=1;for(;y;y>>=1,x=1ll*x*x%M)if(y&1)z=1ll*z*x%M;return z;}
//struct node{int to,ne}e[];
//void add(int x,int y){e[++tot]=(node){y,h[x]},h[x]=tot;}
int n,i,a[N],sum,k;
signed main(){
	scanf("%d",&n);
	for (i=1;i<=n;i++) scanf("%d",&a[i]),sum+=abs(a[i]);
	sort(a+1,a+n+1);
	for (k=1;a[k]<0;k++);
	if (k==n+1 || k==1) sum-=2*abs(k==1?a[1]:a[n]);
	if (k>1) k--;
	printf("%d\n",sum);
	for (i=1;i<min(k,n-1);i++) printf("%d %d\n",a[n],a[i]),a[n]-=a[i];
	for (i=k;i<n-1;i++) printf("%d %d\n",a[i],a[i+1]),a[i+1]=a[i]-a[i+1];
	printf("%d %d",a[n],a[n-1]);
}