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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>

long long kai(long long n,long long j){
    long long k=1,h=1;
    long long t;
    if(n>j){
        t=j;
        j=n;
        n=t;
    }
    for(int i=j+1;i<=n+j;i++){
        k*=i;
        k=k%1000000007;
    }
    if(n!=0){
        for(int i=1;i<=n;i++){
            h*=i;
            h=h%1000000007;
        }
    }
    k=k/h;
    k=k%1000000007;
    return k;
}

int main(){
    long long a,b=0,n,m,o=0,p=1,s,t,u;
    scanf("%lld %lld",&n,&m);
    if(m>n/2){
        printf("0\n");
        exit(0);
    }
    for(int i=0;i<m;i++){
        scanf("%lld",&a);
        if(a==b){
            printf("0\n");
            exit(0);
        }
        a=a-1;
        s=(a-b)/2;
        t=(a-b)%2;
        for(long long i=s;i>=0;i--){
            u=t+2*(s-i);
            o+=kai(i,u);
        }
        p*=o;
        p=p%1000000007;
        o=0;
        b=a+2;
    }
    s=(n-b)/2;
    t=(n-b)%2;
    for(long long i=s;i>=0;i--){
        u=t+2*(s-i);
        o+=kai(i,u);
    }
    p*=o;
    p=p%1000000007;
    printf("%lld\n",p);
    return 0;
}