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();
		boolean impo = false;
		long a[] = new long[m + 1];
		long b[] = new long[m + 2];
		a[0] = -9;
		for (int i = 1; i <= m; i++) {
			a[i] = sc.nextInt();
			b[i] = a[i] - a[i - 1] - 1;
			if (a[i] - a[i - 1] == 1) {
				impo = true;
			}
		}

		b[1] = a[1];
		b[m + 1] = n - a[m];



		if(impo) {
			System.out.println(0);
		} else {

			long pat = 0;
			long ans = 1;
			long nidan = 0;
			long idan = 0;

			long kni = 0;
			long kic = 0;
			for (int i = 1; i <= m + 1; i++) {
				nidan = (b[i] - 1) / 2;
				idan = (b[i] - 1) % 2;
				pat = 0;
				//System.out.println();
				for (long k = 0; k <= nidan; k++) {
					kni = nidan - k;// 二段をこれだけ
					kic = (b[i] - 1) - kni * 2;

				//	System.out.println(kni + " +" + kic);
				//System.out.println(con(kni + kic, Math.min(kni, kic)));
					pat = pat + con(kni + kic, kic) % 1000000007;
					pat = pat % 1000000007;
				}
				// System.out.println(b[i]);
				//System.out.println("!" + pat);
				//System.out.println((ans * pat) % 1000000007);
				ans = (ans * pat) % 1000000007;
			}

			System.out.println(ans);
		}

	}

	static public long con(long n, long r) {
		long num = 1;
		for (long i = 1; i <= r; i++) {
			num = num * (n - i + 1) / i;
		}
		return num;
	}
}