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 <iostream>
#include <vector>
#include <queue>
#include <algorithm>

using namespace std;

typedef long long ll;

int main() {
    int n, k = 0, i = 0;  cin >> n;
    vector<ll> a(n), x(n-1), y(n-1);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    sort(a.begin(), a.end());
    while (k+1 < n-1 && a[k+1] < 0)   k++;
    ll ans = 0, p = a[n-1], m = a[0];
    for (int j = 1; j <= k; j++) {
        x[i] = p;   y[i] = a[j];
        p -= a[j];  i++;
    }
    for (int j = k+1; j < n-1; j++) {
        x[i] = m;   y[i] = a[j];
        m -= a[j];  i++;
    }
    x[i] = p;   y[i] = m;
    cout << p - m << endl;
    for (int j = 0; j < n-1; j++)
        cout << x[j] << " " << y[j] << endl;
    return 0;
}