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: Time Limit Exceeded)

#include<stdio.h>
#include<stdlib.h>
int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    int a[m];
    for(int i = 0;i < m;i++){
        scanf("%d",&a[i]);
    }
    int dp[n + 1];
    for(int i = 0;i < (n + 1);i++){
        dp[i] = 0;
    }
    dp[0] = 1;
    for(int i = 1;i < (n + 1);i++){
        
        if(i == 1){
            dp[i] = 1;
            }else{
            dp[i] += dp[i - 1];
            dp[i] += dp[i - 2];
            }
        if(dp[i] > 1000000007){
            dp[i] %= 1000000007;
        }    


        for(int j = 0;j < m;j++){
            if(i == a[j]){
                dp[i] = 0;
            }
        }
    }
    printf("%d",dp[n]);
    return 0;
}