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(auto &x: a) cin >> x;

    sort(a.begin(), a.end());

    long ma = a[n-1], mi = a[0];

    long r = ma - mi;
    for(int i=1; i<n-1; i++) {
        r += max(a[i], -a[i]);
    }
    cout << r << endl;

    for(int i=1; i<n-1; i++) {
        long aa = a[i];
        if(aa > 0) {
            cout << mi << ' ' << aa << endl;
            mi -= aa;
        } else {
            cout << ma << ' ' << aa << endl;
            ma -= aa;
        }
    }
    cout << ma << ' '<< mi << endl;
}
