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: Wrong Answer)

#include<stdio.h>
#include<cstdio>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string>
#include<cmath>
#include<vector>

#include<stack>
#include<queue>
#include<time.h>
#include<set>
#include<map>
#include <iomanip>
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define REP(i,a,b) for(int i=a;i>b;i--)
#define vint vector<int> 
#define vvint vector<vint>
#define ct(a) cout<<a<<endl
using namespace std;
typedef long long ll;
const ll MOD = 1e9 + 7;
const ll mod = 998244353;


string S;

ll ans=1,N,M,A[100005];
queue<int>que;
ll fib[100005];
int main() {

	cin >> N >> M;
	FOR(i, 1, M+1)cin >> A[i];
	A[0] = -1;
	fib[1] = 1; fib[2] = 1;
	FOR(i, 3, 100002) {
		fib[i] = fib[i - 1] + fib[i - 2];
	}
	FOR(i, 1, M+1) {
		if (A[i] - A[i - 1] == 1) { ans = 0; break; }
		ans = (ans*fib[A[i] - A[i - 1]-1])%MOD;
	}
	if (M > 0) {
		ans = (ans*fib[N - A[M]])%MOD;
	}

	cout << ans << endl;
	return 0;
}