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.util.*;

public class Main {

    public static void main(String[] args) {
//        long startTime = System.currentTimeMillis();
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int M = scanner.nextInt();
        long[] nValue = new long[N + 1];
        nValue[0] = 1L;
        for (int i = 0; i < M; i++) {
            int a = scanner.nextInt();
            nValue[a] = -1;
        }
        scanner.close();

        long calcValue = 1000000007;

        for (int i = 0; i < N + 1; i++) {
            if (nValue[i] < 0) {
                continue;
            }
            // 1段上がる場合
            int one = i + 1;
            if (one <= N && nValue[one] >= 0) {
                nValue[one] = (nValue[i] + nValue[one]) % calcValue;
            }
            // 2段上がる場合
            int two = i + 2;
            if (two <= N && nValue[two] >= 0) {
                nValue[two] = (nValue[i] + nValue[two]) % calcValue;
            }
        }

        System.out.println(nValue[N]);
    }
}