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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)

    val N, M = sc.nextInt()
    val A = Array.fill(M)(sc.nextInt()).toSet

    // dp(i) := i段目にたどり着くための移動方法が何通りか？
    // 最終的にdp(N) を求める
    var dpi_a = 1L
    var dpi_b = 0L

    if (A.contains(1))
      dpi_b = 0L
    else
      dpi_b = dpi_a

    (1 until N) foreach { i =>
      if (A.contains(i+1)) {
        // 壊れているので0通り
        dpi_a = dpi_b
        dpi_b = 0
      } else { // 壊れていないので, 一個前と二個前のをたす
        val temp = dpi_b
        dpi_b = dpi_a + dpi_b
        dpi_a = temp
      }
    }
    println(dpi_b % 1000000007)
  }

}