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 <iostream>
#include <vector>
#include <algorithm>
#include <utility>

using namespace std;

int main(){
    int n;
    cin >> n;
    vector<long long> a(n);
    for(auto&& i : a){
        cin >> i;
    }
    sort(begin(a), end(a));
    vector<pair<long long, long long>> ans;
    int i = 1;
    for(; i < n - 1 && a[i] <= 0; i++){
        ans.push_back(make_pair(a[n - 1], a[i]));
        a[n - 1] -= a[i];
    }
    for(; i < n - 1; i++){
        ans.push_back(make_pair(a[0], a[i]));
        a[0] -= a[i];
    }
    cout << a[n - 1] - a[0] << "\n";
    for(auto&& i : ans){
        cout << i.first << " " << i.second << "\n";
    }
    cout << a[n - 1] << " " << a[0] << "\n";
}