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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int[] a = new int[m + 2];
		long count = 1L;

		//壊れている場所がa[1]~a[m] a[0]=0(最初の段), a[m+1]=n(最後の段)
		a[0] = 0;
		a[m + 1] = n;

		for (int i = 1; i < m + 1; i++) {
			a[i] = sc.nextInt();
		}

		//x段のぼる場合の数の計算
		//数がデカそうなのでlong型
		long[] x = new long[n];
		x[0] = 1;
		x[1] = 1;
		for (int i = 2; i < n+1; i++) {
			x[i] = x[i - 1] + x[i - 2];
		}

		//2段連続で壊れてたら0通りで終了
		for (int i = 1; i < m; i++) {
			if (a[i + 1] - a[i] == 1) {
				System.out.println("0");
				System.exit(0);
			}
		}

		for (int i = 1; i <= m + 1; i++) {
			if (m == 0) {
				count = x[n];
			} else if (a[i] == 1 || a[i] == n - 1) {
				continue;
			} else if (i == 1 || i == m + 1) {
				count = count * x[a[i] - a[i - 1] - 1];
			} else {
				count = count * x[a[i] - a[i - 1] - 2];
			}
			count %= 1000000007;
		}

		System.out.println(count);

	}
}