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(){

    long int ans = 1;

    int N, M;
    scanf("%d%d", &N, &M);

    int a[M+2];

    a[0] = -1;
    a[M + 1] = N+1;

    for (int i = 1; i < M+1; i++){

        scanf("%d", &a[i]);

        if(a[i] == a[i-1] + 1){
            ans = 0;
            break;
        }
    }

    if(ans == 1){

        long int pattern[N+1];

        pattern[0] = 1;
        pattern[1] = 1;
        pattern[2] = 2;

        for (int i = 3; i < N+1; i++){
            pattern[i] = pattern[i - 1] + pattern[i - 2];
        }

        for (int i = 1; i < M+2; i++){

            ans = (ans * pattern[a[i] - a[i - 1] - 2]) % 1000000007;
        }
    }

    printf("%ld", ans);

    return 0;
}