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 res = A[N - 1] - A[0];
    for (int i = 1; i < N - 1; i++) {
        res += abs(A[i]);
    }
    cout << res << endl;

    for (int i = N - 2; i > 0 && A[i] >= 0; i--) {
        cout << A[0] << " " << A[i] << endl;
        A[0] -= A[i];
    }
    for (int i = 1; i < N - 1 && A[i] < 0; i++) {
        cout << A[N - 1] << " " << A[i] << endl;
        A[N - 1] -= A[i];
    }

    cout << A[N - 1] << " " << A[0] << endl;
    return 0;
}