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 oo 2000000000
#define mod 998244353 
const int N = 200010;
int n , arr[N];

vector< pair<int,int> > v;

int main(){
    scanf("%d",&n);
    for(int i = 0 ;i < n;i++){
        scanf("%d",&arr[i]);
    }
    sort(arr,arr+n);
    for(int i = 1;i < n - 1;i++){
        if(arr[i] > 0){
            v.push_back(make_pair(arr[0] , arr[i]));
            arr[0] = arr[0] - arr[i];
        }
        else{
            v.push_back(make_pair(arr[n - 1] , arr[i]));
            arr[n - 1] = arr[n - 1] - arr[i];
        }
    }
    v.push_back(make_pair(arr[n - 1] , arr[0]));
    int cur = arr[n - 1] - arr[0];
    cout << cur << endl;
    for(int i = 0 ;i < (int)v.size();i++){
        printf("%d %d\n",v[i].first,v[i].second);
    }
    return 0;
}