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 (1.3.71) (judge verdict: Accepted)

fun main() {
    abc129c()
}

fun abc129c() {
    val (n, m) = readLine()!!.split(" ").map { it.toInt() }
    val step = LongArray(n + 1) { 0L }
    repeat(m) {
        val i = readLine()!!.toInt()
        step[i] = -1L
    }
    step[0] = 1
    for (i in 1..n) {
        if (step[i] == -1L) continue
        val one = step[i - 1]
        val two = if (i > 1) step[i - 2] else 0L
        var s = 0L
        if (one != -1L) s += one
        if (two != -1L) s += two
        step[i] = s % 1_000_000_007L
    }
    println(step[n].coerceAtLeast(0L))
}