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;

typedef pair <int, int> ii;

const int Maxn = 100005;
const int Inf = 2000000007;

int n;
int a[Maxn];
vector <ii> res;

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    sort(a + 1, a + n + 1);
    int fir = a[1];
    int lst = a[n];
    for (int i = 2; i < n; i++)
        if (a[i] >= 0) {
            res.push_back(ii(fir, a[i]));
            fir -= a[i];
        } else {
            res.push_back(ii(lst, a[i]));
            lst -= a[i];
        }
    res.push_back(ii(lst, fir));
    lst -= fir;
    printf("%d\n", lst);
    for (int i = 0; i < res.size(); i++)
        printf("%d %d\n", res[i].first, res[i].second);
    return 0;
}
