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>

typedef long long int ll;

int main(void){
    int M, N;
    int a[100000];
    int br[100000];
    int i = 0;
    int j = 0;
    ll ans = 1;
    ll div=1000000007;
    ll st[100001];

	for(i = 0; i < 100000; i++){
    	a[i] = 0;
    }

  
    scanf("%d %d", &N, &M);
    for(i = 0; i < M; i++){
        scanf("%d", &a[i]);
        if(i != 0 && a[i-1]+1 == a[i]){
            printf("0");
            return 0;
        }
    }

    br[0] = 0;

    for(i = 1; i <= N; i++){
        if(i == a[j]){
            br[i] = 1;
          	j++;
        }else{
            br[i] = 0;
        }
    }

    st[0] = 1;
    if(br[1] == 0){
        st[1] = 1;
    }else{
        st[2] = 2;
    }

    for(i = 2; i <= N; i++){
        if(br[i] == 0){
            if(br[i-1] == 0 && br[i-2] == 0){
                st[i] = st[i-1] + st[i-2];
            }else if(br[i-1] == 1){
                st[i] = st[i-2];
            }else if(br[i-2] == 1){
                st[i] = st[i-1];
            }
          	st[i] = st[i]%div;
            ans = st[i];
        }
    }


    printf("%lld", ans);
    return 0;
}