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<limits.h>


int main(void){
    long long int n, m, a[100005]={}, b[100005]={}, l=0, s=1;
    scanf("%lld %lld", &n, &m);
    a[0]=-1;
    for(int i=1; i<=m; i++){
        scanf("%lld", &a[i]);
    }
    a[m+1]=n+1;
    for(int i=0; i<=m; i++){
        a[i]=a[i+1]-a[i]-1;
    }
    for(int i=0; i<=m; i++){
        if(a[i]>l){
            l=a[i];
        }
    }
    b[0]=0;
    b[1]=1;
    for(int i=2; i<=l; i++){
        b[i]=b[i-1]+b[i-2];
    }
    // for(int i=0; i<=l; i++){
    //     printf("%d\n", b[i]);
    // }
    for(int i=0; i<=m; i++){
        s=(s*(b[a[i]]%1000000007))%1000000007;
    }
    printf("%lld", s);
    return 0;
}