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>

int main(void){
    int n, m;
    scanf("%d %d",&n ,&m);
    int a[m];
    long long int b[n];
    int i, j, k=0;
    int flag;
    long long int ans;
    for(i=0;i<m;i++){
        scanf("%d",&a[i]);
    }

    // shokichi
    if(a[0] == 1 && a[1] == 2){
        printf("0");
        return 0;
    }
    if(a[0] == 2){
        b[0] = 1;
        b[1] = 0;
    }
    else if(a[0] == 1 && a[1] != 2){
        b[0] = 0;
        b[1] = 1;
    }
    else{
        b[0] = 1;
        b[1] = 2;
    }

    //keisan
    for(i=0;i<n-2;i++){
        b[i+2] = b[i+1] + b[i];
        for(j=k;j<m;j++){
            if(a[j] == i+3){
                flag++;
                k = j;
                break;
            }
        }
        if(flag == 1){
            b[i+2] = 0;
            flag--;
        }
    }
    ans = b[n-1]%1000000007;
    printf("%lld\n",ans);
    return 0;
    
}