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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<stdbool.h>
#include<limits.h>
#include<float.h>

#define NUM 100001

int step(int n, int i, int m, int *ans){
	if(i==-1) return 0;
	else if (i==0) return 1;
	else if (ans[i] != -1) return ans[i];
	else return ans[i] = (step(n, i - 1, m, ans) + step(n, i - 2, m, ans)) % 1000000007;
}

int main(){
	int n, m, ans[NUM], a;
	scanf("%d%d\n", &n, &m);
	for(int i=0; i<n+1; i++){
		ans[i] = -1;
	}
	for(int i=0; i<m; i++){
		scanf("%d", &a);
		ans[a] = 0;
	}
	printf("%d\n", step(n, n, m, ans));
	return 0;
}