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 <cstdio>
#include <algorithm>
// C - Successive Subtraction

using namespace std;

int A[100005];

int main() {
    int N,i,M,lst,cur1,cur2;

    scanf("%d",&N);

    for (i = 0;i < N;i++)
        scanf("%d",&A[i]);

    sort(A,A + N);

    M = A[N - 1] - A[0];
    lst = 0;
    for (i = 1;i < N - 1;i++) {
        if (A[i] < 0) {
            lst = i;
            M -= A[i];
        }
        else
            M += A[i];
    }

    printf("%d\n",M);

    cur1 = A[lst];
    for (i = lst + 1;i < N - 1;i++) {
        printf("%d %d\n",cur1,A[i]);
        cur1 -= A[i];
    }

    cur2 = A[N - 1];
    for (i = 0;i < lst;i++) {
        printf("%d %d\n",cur2,A[i]);
        cur2 -= A[i];
    }

    printf("%d %d\n",cur2,cur1);
    
    return 0;
}
