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 <algorithm>
#include <vector>
using namespace std;

int N;

int main() {
    cin >> N;
    vector<long long> A(N);
    for(int i = 0; i < N; i++) cin >> A[i];
    sort(A.rbegin(), A.rend());
    vector<pair<long long, long long>> S;

    for(int i = 1; i < N-1; i++) {
        if(A[i] >= 0) {
            S.push_back(make_pair(A[N-1], A[i]));
            A[N-1] -= A[i];
        } else {
            S.push_back(make_pair(A[0], A[i]));
            A[0] -= A[i];
        }
    }
    S.push_back(make_pair(A[0], A[N-1]));
    A[0] -= A[N-1];
    cout << A[0] << endl;
    for(pair<int, int> p: S) {
        cout << p.first << " " << p.second << endl;
    }
}