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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val sc = Scanner(System.`in`)
    val n = sc.nextInt()
    val m = sc.nextInt()
    val a = (0 until m).map { 
        sc.nextInt()
     }

     abc129c(n, a)
}

fun abc129c(n: Int, a: List<Int>) {
    val safe = BooleanArray(n + 1) { true }
    for (ai in a) {
        safe[ai] = false
    }
    val dp = LongArray(n + 1) { 0 }
    dp[0] = 1L
    if (safe[1]) dp[1] = 1L
    for (i in 2..n) {
        if (safe[i - 2]) dp[i] += dp[i - 2]
        if (safe[i - 1]) dp[i] += dp[i - 1]
    } 
    val mod: Long = (Math.pow(10.0, 9.0) + 7).toLong()
    val ans: Long = dp[n] % mod
    println(ans)
}