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;
    vector<long long int> V;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int tmp;
        cin >> tmp;
        V.push_back(tmp);
    }
    sort(V.begin(), V.end());
    long long int ans = 0;
    for (int i = 0; i < n; i++) {
        if (i == 0)
            ans -= V[i];
        else if (i == n - 1)
            ans += V[i];
        else
            ans += abs(V[i]);
    }
    cout << ans << "\n";
    for (int i = 1; i < n - 1; i++) {
        if (V[i] >= 0) {
            cout << V[0] << " " << V[i] << "\n";
            V[0] -= V[i];
        }
        else {
            cout << V[n - 1] << " " << V[i] << "\n";
            V[n - 1] -= V[i];
        }
    }
    cout << V[n - 1] << " " << V[0] << "\n";
}
