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: Accepted)

fun main(args: Array<String>){
    val (N, M) = readLine()!!.split(" ").map { it.toInt() }
    val A = IntArray(M){ readLine()!!.toInt() }
    val mod = 1000000007L

    val dp = LongArray(N+1)
    dp[0] = 1
    if(!A.contains(1)){
        dp[1] = 1
    }
    A.forEach {
        dp[it] = -1L
    }
    for(i in 2..N){
        if(dp[i] != -1L){
            dp[i] = (if(dp[i - 1] != -1L) dp[i - 1] else 0) + (if(dp[i - 2] != -1L) dp[i - 2] else 0)
            dp[i] = dp[i] % mod
        }
    }
    println(dp[N])
}