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)

fun main(args: Array<String>) {
    val (N, M) = readLine()!!.split(" ").map { it.toInt() }
    var brokenStairs = mutableSetOf<Int>()
    var memo = mutableMapOf<Int, Long>()
    for (i in 1..M) {
        brokenStairs.add(readLine()!!.toInt())
    }
    print(fibonacci(N, memo, brokenStairs) % 1000000007)
}

fun fibonacci(n: Int, memo: MutableMap<Int, Long>, brokenStairs: Set<Int>): Long {
    if (brokenStairs.contains(n - 1)) {
        fibonacci(n - 2, memo, brokenStairs)
    }
    if (memo.containsKey(n)) {
        return memo[n]!!
    }
    when (n) {
        0 -> memo[n] = 1
        else -> {
            if (brokenStairs.contains(n)) {
                memo[n] = 0
                fibonacci(n - 1, memo, brokenStairs)
            } else {
                memo[n] = fibonacci(n - 1, memo, brokenStairs) + fibonacci(n - 2, memo, brokenStairs)
            }
        }
    }
    return memo[n]!!
}