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: Accepted)

import java.math.BigInteger;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class Main {
	Integer steps[];
	Map<Integer, BigInteger> map;

	public Main(int n, Integer a[]) {
		steps = new Integer[n];
		map = new TreeMap<>();
		for (int i = 0; i < steps.length; i++) {
			steps[i] = 1;
		}
		for (int e : a) {
			steps[e] = 0;
		}

	}

	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int m = s.nextInt();
		Integer a[] = new Integer[m];
		for (int i = 0; i < m; i++) {
			a[i] = s.nextInt();
		}
		System.out.println(
				new Main(n, a).calc(0).mod(new BigInteger("1000000007")));
	}

	public BigInteger calc(int prefix) {
		BigInteger shortcut = map.get(prefix);
		BigInteger result;
		if (shortcut != null) {
			return shortcut;
		}
		if (prefix >= steps.length) {
			result = BigInteger.ZERO;
		} else if (prefix == steps.length - 1) {
			result = BigInteger.ONE;
		} else if (prefix == steps.length - 2) {
			result = BigInteger.ONE
					.add(((steps[prefix + 1] == 1) ? BigInteger.ONE
							: BigInteger.ZERO));
		} else {
			result = ((steps[prefix + 1] == 1) ? calc(prefix + 1)
					: BigInteger.ZERO)
							.add((steps[prefix + 2] == 1) ? calc(prefix + 2)
									: BigInteger.ZERO);
		}
		map.put(prefix, result);
		return result;
	}
}