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 <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;

typedef long long ll;

ll n;
ll a[100005];
ll ans;

int main()
{
    scanf("%lld", &n);
    for(ll i = 0; i < n; i++)
    {
        scanf("%lld", &a[i]);
    }
    sort(a, a + n);
    for(ll i = 1; i < n - 1; i++)
    {
        ans += abs(a[i]);
    }
    ans += a[n - 1] - a[0];
    printf("%lld\n", ans);
    for(ll i = 1; i < n - 1; i++)
    {
        if(a[i] > 0) printf("%lld %lld\n", a[0], a[i]), a[0] -= a[i];
        else printf("%lld %lld\n", a[n - 1], a[i]), a[n - 1] -= a[i];
    }
    printf("%lld %lld\n", a[n - 1], a[0]);
    return 0;
}
