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: Accepted)

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


int main(void){
    int n,m,i,k=0;
    scanf("%d%d",&n,&m);
    int a[m],p[n];
    /*m段目までの行き方p[m-1]*/
    for(i=0;i<m;i++){
        scanf("%d",&a[i]);
    }
    for(i=0;i<n;i++){
        p[i] = 0;
    }

    for(i=0;i<n;i++){

        if(i==0){
            if(i==a[k]-1){
            k++;
            continue;
            }
            p[i]=1;
            // printf("%d\n",p[i]);
            continue;
        }

        if(i==1){
            if(i==a[k]-1){
            k++;
            continue;
            }
            p[i]=p[i-1]+1;
            //printf("%d\n",p[i]);
            continue;
        }

        if(i==a[k]-1){
            k++;
            continue;
        }

        p[i] = (p[i-1] + p[i-2]) % 1000000007;
        // printf("%d\n",p[i]);

    }
    printf("%d\n",p[n-1]);
    return 0;
}