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;

int main()
{
    int N;
    cin >> N;
    vector<int> A(N);
    for(int i=0;i<N;i++){
        cin >> A[i];
    }
    sort(A.begin(), A.end());
    int i=1;
    vector<pair<int, int>> res;
    for(; i<N-1; i++){
        if(A.at(i) > 0) break;
        res.emplace_back(A.back(), A.at(i));
        A.back() -= A.at(i);
    }
    for(; i<N-1; i++){
        res.emplace_back(A.front(), A.at(i));
        A.front() -= A.at(i);
    }
    res.emplace_back(A.back(), A.front());

    cout << (res.back().first - res.back().second) << endl;
    for(auto v : res){
        cout << v.first << " " << v.second << endl;
    }

    return 0;
}