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(void){
    int mod = 1000000007;
    int N, M;
    scanf("%d%d", &N, &M);
    long fib[N+1]; //fib[i] は i+1段の段数がある時の上りの数
    fib[0] = 0;
    fib[1] = 1;
    fib[2] = 1;
    for (int i = 3; i < N+1; i++){
        fib[i] = (fib[i-2] + fib[i-1]) % mod;
    }
    
    int a[M+1];
    a[0] = -1;
    long ans = 1;
    for (int i = 1; i < M+1; i++){
        scanf("%d", &a[i]);
        if(a[i] - a[i-1] == 1){
            printf("%d\n", 0);
            return 0;
        }
    }
    for (int i = 1; i <= M; i++){
        ans *= fib[(a[i] - a[i-1]) - 1];
        ans = ans%mod;
    }
    ans *= fib[N-a[M]];
    ans = ans%mod;
    printf("%ld\n", ans);
    return 0; 
}