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: Wrong Answer)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		final int D = 1000000007;
		Scanner in = new Scanner(System.in);

		int N = in.nextInt();
		int M = in.nextInt();
		int[] a = new int[M + 2];
		a[0] = -1;
		a[M + 1] = N + 1;

		// 壊れ段が2連続 → 答え0個
		boolean enable = false;
		for ( int i = 1; i <= M; i++ ) {
			a[i] = in.nextInt();
			if ( a[i] == a[i - 1] + 1 ) {
				enable = true;
			}
		}
		if ( enable ) {
			System.out.println(0);
			in.close();
			return;
		}

		long result = 1;
		// 壊れ段ごとに区切って見ていく
		for ( int i = 0; i <= M; i++ ) {
			int steps = (a[i + 1] - 1) - (a[i] + 1); // 壊れ段の直後から直前までの段数
			long temp = 0;

			// one_step：「一段上る」の回数
			for ( int one_step = steps % 2; one_step <= steps; one_step += 2 ) {
				// option：「一段上る」＋「二段上る」
				int option = (steps - one_step) / 2 + one_step;
				temp += nCr(option, one_step);
			}
			result = (result * (temp % D)) % D;
			temp = 0;
		}
		System.out.println(result);
		in.close();
	}

	//組み合わせの数nCrを計算
	private static long nCr(int n, int r) {
		long num = 1;
		for ( int i = 1; i <= r; i++ ) {
			num = num * (n - i + 1) / i;
		}
		return num;
	}
}