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>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define PI 3.141592653589793

int main(){
	int n, m;
	scanf("%d%d", &n, &m);
	int broken[m+1];
	for(int i=0; i<m; i++){
		scanf("%d", &broken[i]);
	}
	
	int i=3, j=0;
	unsigned long int temp, way1, way2;
	if(broken[0] == 1){
		if(broken[1] == 2){
			way1 = 0;
			i = n+1;
			j = 2;
		}
		else{
			way1 = 1;
			way2 = 0;
			j = 1;
		}
	}
	else{
		if(broken[0] == 2){
			way1 = 0;
			way2 = 1;
			j = 1;
		}
		else{
			way1 = 2;
			way2 = 1;
		}
	}
	while(i<=n){
		if(i == broken[j]){
			temp = 0;
			j++;
			//printf("Yes");
		}
		else{
			temp = way1+way2;
		}
		way2 = way1;
		way1 = temp;
		//printf("%d段目:%lu\n", i, temp);
		temp = temp%1000000007;
		way1 = way1%1000000007;
		way2 = way2%1000000007;
		i++;
	}
	
	way1 = way1%1000000007;
	printf("%lu", way1);
	
	
	return 0;
}