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;
using Int = long long;
int main()
{
    cin.tie(nullptr); ios::sync_with_stdio(false);
    int N; cin >> N;
    vector<Int> A(N); for (auto &a : A) cin >> a;
    sort(begin(A), end(A));
    vector<bool> neg(N);
    for (int i = 0; i < N; i++) neg[i] = A[i] < 0 ? true : false;
    neg[0] = true;
    neg[N-1] = false;
    Int ans = 0;
    for (int i = 0; i < N; i++) ans += (neg[i] ? -1 : +1) * A[i];
    cout << ans << '\n';
    for (int i = 1; i < N-1; i++) if (!neg[i]) {
        cout << A[0] << ' ' << A[i] << '\n';
        A[0] -= A[i];
    }
    for (int i = 0; i < N-1; i++) if (neg[i]) {
        cout << A[N-1] << ' ' << A[i] << '\n';
        A[N-1] -= A[i];
    }
    return 0;
}
