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>
#include <stdlib.h>

int main(int argc, const char * argv[]) {
    int N, M;
    int i, j;
    int *aM;
    int hole = 0;
    long long int *caseNum;
    
    scanf("%d",&N);
    scanf("%d",&M);
    aM = (int *)calloc(M, sizeof(int));
    caseNum = (long long int *)calloc(N, sizeof(long long int));
    
    for(i=0; i<M; i++){
        scanf("%d",&aM[i]);
    }
    
    caseNum[0] = 1;
    caseNum[1] = 2;
    for (j=0; j<M; j++) {
        if(aM[j] == 1){
            caseNum[0] = 0;
            caseNum[1] = 1;
            hole++;
        }
        if(aM[j] == 2){
            caseNum[1] = 0;
            hole++;
            break;
        }
        if(aM[j] > 2){break;}
    }
    
    for(i=2; i<N; i++){
        if(aM[hole] == i+1){
            caseNum[i] = 0;
            hole++;
        }else{
            caseNum[i] = (caseNum[i-1] + caseNum[i-2]) % 1000000007;
        }
    }
    
    printf("%lld\n", caseNum[N-1]);
    
    free(aM);
    free(caseNum);
    return 0;
}