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.math.BigInteger;
import java.util.*;
import java.util.stream.IntStream;

public class Main {

    private static final BigInteger ZERO = BigInteger.ZERO;
    private static final BigInteger ONE = BigInteger.ONE;
    private static final BigInteger TWO = ONE.add(ONE);

    private static final Map<BigInteger, BigInteger> FIB_MAP = new HashMap<>();
    static {
        FIB_MAP.put(ZERO, ZERO);
        FIB_MAP.put(ONE, ONE);
    }

    public static BigInteger fibo(BigInteger num){
        if (num==BigInteger.valueOf(-1)) return BigInteger.ZERO;
        if (num==BigInteger.ZERO) return BigInteger.ONE;
        if (num==BigInteger.ONE) return BigInteger.ONE;
        return FIB_MAP.computeIfAbsent(num, key -> fibo(num.subtract(ONE)).add(fibo(num.subtract(TWO))));
//        if (num==-1){
//            return 0;
//        }
//        if (num==0) {
//            return 1;
//        }
//        if (num==1) {
//            return 1;
//        }
//        return fibo(num-1) + fibo(num-2);
    }

    public static void main(String args[]) {


        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int m = sc.nextInt();
        BigInteger a[] = new BigInteger[m];

        IntStream.range(1, n).forEach(i -> fibo(BigInteger.valueOf(i)));

        BigInteger sum = BigInteger.valueOf(1);
        BigInteger x = BigInteger.valueOf(1000000007);

        for (int i=0; i<m; i++){
            a[i] = sc.nextBigInteger();
        }

        BigInteger multi[] = new BigInteger[m+1];

        multi[0] = fibo(a[0].subtract(ONE)).remainder(x);
        multi[m] = fibo(BigInteger.valueOf(n).subtract(a[m-1]).subtract(ONE)).remainder(x);

        for (int j=1; j<m; j++){
            multi[j] = fibo(a[j].subtract(a[j-1]).subtract(TWO)).remainder(x);
        }

        for (int k=0; k<m+1; k++) {
            sum = sum.multiply(multi[k]);
        }

        System.out.println(sum.remainder(x));

    }
}