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 long D = 1000000007;
		long result = 1;

		Scanner in = new Scanner(System.in);
		int N = in.nextInt();
		int M = in.nextInt();

		// 壊れ段 スタート直前とゴール直後も登録しておく
		int[] a = new int[M + 2];
		a[0] = -1;
		a[M + 1] = N + 1;

		for ( int i = 1; i <= M; i++ ) {
			a[i] = in.nextInt();
			result = (result * getStepPatterns(a[i] - a[i - 1] - 1)) % D;
		}
		result = (result * getStepPatterns(a[M + 1] - a[M] - 1)) % D;

		System.out.println(result);
		in.close();
	}

	// n段上のステップパターンを計算する（nにはスタートとゴールの段を含む）
	private static long getStepPatterns(int ｎ) {
		long D = 1000000007;
		long ptrn = 0;

		if ( ｎ <= 0 ) ptrn = 0;
		else if ( ｎ == 1 || ｎ == 2 ) ptrn = 1;
		else
		// (p+2)段目のパターン = (p+1)段目のパターン + p段目のパターン
		if ( 3 <= ｎ ) {
			long p0, p1, p2;
			p0 = p1 = p2 = 1;

			for ( int i = 2; i < ｎ; i++ ) {
				p2 = p1 + p0;
				if ( p2 > D ) p2 %= D;
			}
			ptrn = p2;
		}
		return ptrn;
	}
}