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)

using namespace std;
#include<iostream>
#include<vector>
int N,M;
const int DIM=100010;
long X[DIM]; 
int ans=1;
const int base=1000000007; 


int FIB(int i){
    if (X[i] != 0) {return X[i];}
    X[i] = FIB(i-2);
    X[i] = (X[i] + FIB(i-1))% base; 
    return X[i]; 
}


int main(){
    int a,i,j,k,x,y,z;
    int c = 0; 
    long s; 

    cin >> N >> M; 
    // cout << N << " " << M << "\n"; 
    for(i=0;i<DIM;i++){X[i]=0;}
    X[0]=0;
    X[1]=1; 
    X[2]=1;
    for(i=0;i<M;i++){
        cin >> a; 
        if (c == a || a >= N){ 
            cout << 0 << endl; // Exit. Serial skip 
            exit(0);
        }
        s = FIB(a-c);
        ans=(int)(((long)ans * (long)s ) % (long) base);
        // cout << a << " " << s << " " << ans << endl;
        c = a + 1;
    }
    s = FIB(N-c+1);
    ans=(int)(((long)ans * (long)s ) % (long) base);
    // cout << M << " " << s << " " << ans << endl;
    cout << ans  << endl; // Exit. Serial skip 

}