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 {
	static long[][] cmb;
	public static void main(String[] args) {
		long mod = 1_000_000_007L;
		Scanner in = new Scanner(System.in);
		int num = in.nextInt();
		int bstepNum = in.nextInt();
		cmb = new long[num + 1][(num + 1)/2];
		int[] brokenStep = new int[bstepNum];
		long ans = 0;
		for (int i = 0; i < bstepNum; i++) {
			brokenStep[i] = Integer.parseInt(in.next());
			if (i != 0 && brokenStep[i-1] + 1 == brokenStep[i]) {
				System.out.println(ans);
				in.close();
				return;
			}
		}
		in.close();
		try {
			ans = 1;
			int pos = 0;
			for (int i = 0; i < bstepNum; i++) {
				int step = brokenStep[i] - pos - 1;
				long tmp = 0;
				for (int j = step, skip = 0; j >= 0; j -= 2, skip++) {
					tmp += combination(j + skip, skip, mod);
				}
				ans *= tmp;
				if (ans > mod) {
					ans %= mod;
				}
				pos = brokenStep[i] + 1;
			}
			long lt = 0;
			for (int i = num - pos, skip = 0; i >= 0; i -= 2, skip++) {
				lt += combination(i + skip, skip, mod);
			}
			ans *= lt;
			ans %= mod;
			System.out.println(ans);
		} catch (ArithmeticException e) {
			System.out.println("0");
		}
	}
	
	public static long combination(int a, int b, long mod) {
		if (cmb[a][b] != 0) return cmb[a][b];
		if (a == b) {
			return cmb[a][b] = 1L;
		} else if (b == 0) {
			return cmb[a][b] = 1L;
		} else if (b == 1) {
			return cmb[a][b] = a;
		} else {
			long tmp = combination(a-1, b-1, mod) + combination(a-1, b, mod);
			if (mod < tmp) {
				tmp %= mod;
			}
			return cmb[a][b] = tmp;
		}
	}
}