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
N = int(input())
A = [int(i) for i in input().split()]
A.sort(reverse=True)

log = []

# すべて正
if A[-1] > 0 :
    a = A.pop()
    b = A.pop()
    log.append([a, b])
    A.append(a - b)
# すべて負
elif A[0] < 0 :
    a = A.pop(0)
    b = A.pop()
    log.append([a, b])
    A = [a-b] + A
    
i = 1
while A[i] > 0 :
    a = A.pop()
    log.append([a, A[i]])
    A.append(a-A[i])
    i += 1
    
while i < len(A) :
    log.append([A[0], A[i]])
    A[0] -= A[i]
    i += 1
    
print(A[0])
for l in log :
    print(l[0], l[1])
    