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: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        // write your code here
        Scanner sc = new Scanner(System.in);

        // abc 129 C

        int n = sc.nextInt();
        int m = sc.nextInt();

        int[] brokenSteps = new int[m];

        if (m > 0) {
            brokenSteps[0] = sc.nextInt();
        }

        for (int i = 1; i < m; i++) {
            brokenSteps[i] = sc.nextInt();
            if (brokenSteps[i] - brokenSteps[i - 1] == 1) {
                // 壊れている段が連続している場合は最上段に到達できない
                System.out.println(0);
                return;
            }
        }

        // 壊れている段を跨ぐ場合は全ての場合で2固定なので検討不要
        // 検討すべき段数をまず出し、その中で含まれる可能性のある最大の2の個数とその場合含まれる1の個数でありえる組み合わせが出せる
        // 後は2の個数を一つずつ減らしてその場合有りえる組み合わせを計算して足していく
        // 真面目にやってると遅すぎて間に合わなそうなので計算の省略を考える
        long ans = 1;
        int start = 0;
        int mod = 1000000007;
        list = new long[n / 2 + 1];
        list[0] = 1;
        for (int i = 1; i <= n / 2; i++) {
            list[i] = list[i - 1] * pow(i,mod-2,mod) % mod;
        }

        for (int i = 0; i < m; i++) {
            int end = brokenSteps[i] - 1;
            if (end > start) {
                int valid = end - start;
                int max2 = valid / 2;
                long pattern = 0;
                for (int j = 0; j <= max2; j++) {
                    int mother = valid - j;
                    pattern += nCk(mother, j, mod);
                    pattern %= mod;
                }
                ans *= pattern;
                ans %= mod;
            }
            start = end + 2;
        }

        int lastStart = m > 0 ? brokenSteps[m - 1] + 1 : 0;
        int valid = n - lastStart;
        int max2 = valid / 2;
        long pattern = 0;
        for (int j = 0; j <= max2; j++) {
            int mother = valid - j;
            pattern += nCk(mother, j, mod);
            pattern %= mod;
        }
        ans *= pattern;
        System.out.println(ans % mod);
    }

    private static long[] list;

    //return nCk mod M (M must be prime number) O(min(k,n-k)*logM)
    public static int nCk(int n,int k,int M) {
        int min = Math.min(k, n-k);
        long ret = list[min];
        for(int i=n-min+1;i<=n;i++) {
            ret = (ret * i) % M;
        }
        return (int)ret;
    }

    //return a^b mod M O(logB)
    public static long pow(long a,long b,int M) {
        long ret = 1;
        long tmp = a;
        while(b>0) {
            if((b&1)==1) ret = (ret * tmp) % M;
            tmp = (tmp * tmp) % M;
            b = b>>1;
        }
        return ret;
    }
}