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.ArrayList

fun main(args: Array<String>) {
    val (n, originalM) = readLine()!!.split(" ").map(String::toInt)
    val originalA = IntArray(originalM)
    repeat(originalM) { i ->
        originalA[i] = readLine()!!.toInt()
    }

    val a = originalA.distinct().toIntArray()
    val m = a.size

    val cache = LongArray(n + 1)

    if (m == 0) {
        println(countPattern(n, cache))
        return
    }

    if (isImpossible(a)) {
        println("0")
        return
    }

    val separated = ArrayList<Int>()
    var prev = a[0]
    separated.add(prev - 1)
    for (i in 1 until m) {
        val curr = a[i]
        separated.add(curr - prev - 2)
        prev = curr
    }
    separated.add(n - prev - 1)
    val filtered = separated.filter { it -> it > 0 }

    var ans: Long = 1
    for (s in filtered) {
        ans *= countPattern(s, cache)
    }
    println(ans % 1000000007)
}

private fun isImpossible(a: IntArray): Boolean {
    for (i in 0 .. a.size - 2) {
        if (a[i] + 1 == a[i + 1]) {
            return true
        }
    }
    return false
}

private fun countPattern(steps: Int, cache: LongArray): Long {
    if (steps < 3) return steps.toLong()

    val cached = cache[steps]
    if (cached != 0L) {
        return cached
    }

    val result = countPattern(steps - 1, cache) + countPattern(steps - 2, cache)
    cache[steps] = result
    return result
}