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>

int main(){
    using namespace std;
    unsigned long N;
    cin >> N;
    vector<long> A(N);
    for(auto& i : A)cin >> i;
    sort(A.rbegin(), A.rend());
    vector<long> sgn(N, 1);
    auto t = min<unsigned long>(N - 1, max(1L, lower_bound(A.rbegin(), A.rend(), 0) - A.rbegin()));
    fill(sgn.rbegin(), sgn.rbegin() + t, -1);
    cout << inner_product(A.begin(), A.end(), sgn.begin(), 0L) << endl;
    for(unsigned long i = 1; i + t < N; ++i){
        cout << A[N - t] << " " << A[i] << endl;
        A[N - t] -= A[i];
    }
    for(unsigned long i = N - t; i < N; ++i){
        cout << A[0] << " " << A[i] << endl;
        A[0] -= A[i];
    }
    return 0;
}