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>
#include<math.h>

int main(void){
    int N,M,i,j,t;
    long long int s = 1;
    scanf("%d %d",&N,&M);
    int a[M+1];
    long long int p[N+2];
    p[0] = 0;
    p[1] = 1;
    a[0] = -1;
    for(i = 2;i<=N+1;i++) p[i] = p[i-2] + p[i-1];
    if(M > 0){
        for(j = 1;j <= M;j++){
            scanf("%d",&a[j]);
            s = s * (p[a[j]-a[j-1]-1]%1000000007) % 1000000007;
        }
        s = s * (p[N-a[M]] % 1000000007) % 1000000007;
    }else{
        s = p[N+1] % 1000000007;
    }
    printf("%lld",s);
    return 0;
}