Problem:
Score : 300 points

Problem Statement

There are N people, conveniently numbered 1 through N.
They were standing in a row yesterday, but now they are unsure of the order in which they were standing.
However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person.
According to their reports, the difference above for person i is A_i.

Based on these reports, find the number of the possible orders in which they were standing.
Since it can be extremely large, print the answer modulo 10^9+7.
Note that the reports may be incorrect and thus there may be no consistent order.
In such a case, print 0.

Constraints

1≦N≦10^5

0≦A_i≦N-1

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the number of the possible orders in which they were standing, modulo 10^9+7.

Sample Input 1

5
2 4 4 0 2

Sample Output 1

4

There are four possible orders, as follows:

2,1,4,5,3

2,5,4,1,3

3,1,4,5,2

3,5,4,1,2

Sample Input 2

7
6 4 0 2 4 0 2

Sample Output 2

0

Any order would be inconsistent with the reports, thus the answer is 0.

Sample Input 3

8
7 5 1 1 7 3 5 3

Sample Output 3

16

Solution in Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

program main
	implicit none
	integer :: N
	integer, allocatable :: A(:)
	integer :: i, ans
	
	read(*, *) N
	allocate(A(1:N+1))
	read(*, *) A(1:N)
	A(N+1) = 1000000
	ans = 1

	call heapsort(N+1,A)
	if(mod(N,2) == 0) then
		do i = 1, N/2
			if(A(2*(i-1)+1) /= A(2*i) .OR. A(2*(i-1)+1) /= 2*i-1) then
				write(*, *) 0
				goto 10
			end if
		end do
		do i = 1, N/2
			ans = mod(ans,1000000007)*2
		end do
		write(*, *) ans
	else
		if(A(1) /= 0) then
			write(*, *) 0
			goto 10
		end if
		do i = 1, (N-1)/2
			if(A(2*i) /= A(2*i+1) .OR. A(2*i) /= 2*i) then
				write(*, *) 0
				goto 10
			end if
		end do
		do i = 1, N/2
			ans = mod(ans,1000000007)*2
		end do
		write(*, *) ans	
	end if
	

	10 continue
	
	contains
	subroutine heapsort(n,array)
		implicit none
		integer,intent(in) :: n
		integer,intent(inout) :: array(1:n)
  
		integer ::i,k,j,l
		double precision :: t
  
		if(n.le.0)then
			write(6,*)"Error, at heapsort"; stop
		endif
		if(n.eq.1)return

		l=n/2+1
		k=n
		do while(k.ne.1)
			if(l.gt.1)then
				l=l-1
				t=array(L)
			else
				t=array(k)
				array(k)=array(1)
				k=k-1
				if(k.eq.1) then
					array(1)=t
					exit
				endif
			endif
			i=l
			j=l+l
			do while(j.le.k)
				if(j.lt.k)then
					if(array(j).lt.array(j+1))j=j+1
				endif
				if (t.lt.array(j))then
					array(i)=array(j)
					i=j
					j=j+j
				else
					j=k+1
				endif
			enddo
			array(i)=t
		enddo

		return
	end subroutine heapsort
end program main