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<iostream>
#include<string>
#include<stack>/*** LIFO *** stack<type> var;
push(), top():to get data, pop():to remove, size(), empty() ***/
using namespace std;
 
#define Fp(I, STAT, END) for(int I = STAT; I < END; I++)
#define Fm(I, STAT, END) for(int I = STAT; I > END; I--)
 
#define MAXVALUE 1000000007
 
int main(){
    int N, M;
    cin >> N >> M;
    int a[M];
    unsigned long ans = 0;
    Fp(i, 0, M) cin >> a[i];
    Fp(i, 0, M - 1){
        if(a[i + 1] == a[i] + 1){
            cout << 0 << endl;
            return 0;
        }
    }
 
    unsigned long b[2] = {};
    int num = 0;
    if(a[0] != 1) b[0] = 1;
    else num++;
    if(a[num] != 2) b[1] = b[0] + 1;
    else num++;
    if(N >= 3){
        Fp(i, 2, N){
            if(a[num] != i + 1){
                ans = b[1] + b[0];
                b[0] = b[1];
                b[1] = ans; 
                //cout << "段数: " << i + 1 << ",ans: " << ans << endl;
            }else{
                b[0] = b[1];
                b[1] = 0;
                num++;
            }
            if(b[1] > MAXVALUE && b[0] > MAXVALUE){
                b[1] -= MAXVALUE;
                b[0] -= MAXVALUE;
            }else if(b[1] == 0 && b[0] > MAXVALUE){
                b[0] -= MAXVALUE;
            }else if(b[0] == 0 && b[1] > MAXVALUE){
                b[1] -= MAXVALUE;
            }
        }
    }else if(N == 2){
        cout << b[1] << endl;
        return 0;
    }
    else{
        cout << b[0] << endl;
        return 0;
    }
 
    cout << ans % MAXVALUE << endl;
    return 0;
}