Problem:
Score : 300 points

Problem Statement

There is a staircase with N steps. Takahashi is now standing at the foot of the stairs, that is, on the 0-th step.
He can climb up one or two steps at a time.

However, the treads of the a_1-th, a_2-th, a_3-th, \ldots, a_M-th steps are broken, so it is dangerous to set foot on those steps.

How many are there to climb up to the top step, that is, the N-th step, without setting foot on the broken steps?
Find the count modulo 1\ 000\ 000\ 007.

Constraints

1 \leq N \leq 10^5

0 \leq M \leq N-1

1 \leq a_1 < a_2 < ... < a_M \leq N-1

Input

Input is given from Standard Input in the following format:

N M
a_1
a_2
.
.
.
a_M

Output

Print the number of ways to climb up the stairs under the condition, modulo 1\ 000\ 000\ 007.

Sample Input 1

6 1
3

Sample Output 1

4

There are four ways to climb up the stairs, as follows:

0 \to 1 \to 2 \to 4 \to 5 \to 6

0 \to 1 \to 2 \to 4 \to 6

0 \to 2 \to 4 \to 5 \to 6

0 \to 2 \to 4 \to 6

Sample Input 2

10 2
4
5

Sample Output 2

0

There may be no way to climb up the stairs without setting foot on the broken steps.

Sample Input 3

100 5
1
23
45
67
89

Sample Output 3

608200469

Be sure to print the count modulo 1\ 000\ 000\ 007.

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

program steps
  implicit none
  integer :: n, m, i, maxstep, ans
  integer,allocatable :: a(:), b(:), fib(:)

  read *, n, m
  allocate(a(m+2),b(m+1))
  a(1) = -1
  do i = 1, m
    read *, a(i+1)
  end do
  a(m+2) = n+1
  do i = 1, m+1
    b(i) = a(i+1) - a(i) - 2
  end do
  maxstep = maxval(b)
  allocate(fib(maxstep+1))
  fib(1) = 1
  fib(2) = 1
  do i = 3, maxstep+1
    fib(i) = mod(fib(i-1)+fib(i-2),1000000007)
  end do
  ans = 1
  do i = 1, m+1
    ans = mod(ans * fib(b(i)+1),1000000007)
  end do
  print '(i0)', ans
end program steps