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 a,b,c,i;
  scanf("%ld %ld",&a,&b);
  c=0;
long int dp[a+1],br[b];
  dp[0]=1;
  
for(i=0;i<b;++i);
  {
  scanf("%ld",&br[i]);
  }
  if(br[c]==1) {dp[1]=0;
                c++;
               }
  else dp[1]=1;
  
for(i=2;i<a+1;++i)
  {if(dp[i-1]==0 && dp[i-2]==0) dp[i]=0;
  
  else if(i==br[c])
     {dp[i]=0;
      c++;
      }
  else{
      dp[i]=(dp[i-1]+dp[i-2])%1000000007;
      }
}

  printf("%ld",dp[a]);

return 0;
}