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>
#define mod 1000000007

int main(){
    long long int dp[100100] = {0},b[100100]={0};
    long long int n,m,i;
    long long int x;

    scanf("%lld %lld",&n,&m);
    for(i = 0 ; i < m ; i++){
        scanf("%lld",&x);
        b[x] = -1;
    }
    
    dp[0] = 1;
    for(i = 1; i <= n ; i++){
       // printf("%d",i);
        
        if(b[i] == -1){
            dp[i] = 0;
        }else{
            if(i == 1){
                dp[i] = 1;
            }else{
             dp[i] = (dp[i-1] + dp[i-2])%mod;
            }
        }
          
        
    }
    printf("%lld\n",dp[n]);

    return 0;
}