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 <bits/stdc++.h>
using namespace std;
typedef long long   signed int LL;
typedef long long unsigned int LU;
#define incID(i, l, r) for(LL i = (l)    ; i <  (r); ++i)
#define incII(i, l, r) for(LL i = (l)    ; i <= (r); ++i)
#define decID(i, l, r) for(LL i = (r) - 1; i >= (l); --i)
#define decII(i, l, r) for(LL i = (r)    ; i >= (l); --i)
#define inc(i, n)  incID(i, 0, n)
#define inc1(i, n) incII(i, 1, n)
#define dec(i, n)  decID(i, 0, n)
#define dec1(i, n) decII(i, 1, n)
#define inID(v, l, r) ((l) <= (v) && (v) <  (r))
#define inII(v, l, r) ((l) <= (v) && (v) <= (r))
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define FI first
#define SE second
#define  ALL(v)  v.begin(),  v.end()
#define RALL(v) v.rbegin(), v.rend()
template<typename T> bool setmin  (T & a, T b) { if(b <  a) { a = b; return true; } else { return false; } }
template<typename T> bool setmax  (T & a, T b) { if(b >  a) { a = b; return true; } else { return false; } }
template<typename T> bool setmineq(T & a, T b) { if(b <= a) { a = b; return true; } else { return false; } }
template<typename T> bool setmaxeq(T & a, T b) { if(b >= a) { a = b; return true; } else { return false; } }
LL mo(LL a, LL b) { assert(b > 0); a %= b; if(a < 0) { a += b; } return a; }
LL fl(LL a, LL b) { assert(b > 0); return (a > 0 ? a / b : (a - b + 1) / b); }
LL ce(LL a, LL b) { assert(b > 0); return (a < 0 ? a / b : (a + b - 1) / b); }
template<typename T> T gcd(T a, T b) { return (b == 0 ? a : gcd(b, a % b)); }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
#define bit(b, i) (((b) >> (i)) & 1)
#define BC __builtin_popcountll
#define SC static_cast
#define SI(v) SC<int>(v.size())
#define SL(v) SC<LL >(v.size())
#define RF(e, v) for(auto & e: v)
#define ef else if
#define UR assert(false)

// ---- ----

int n, m, dp[100001], MOD = 1e9 + 7;

int main() {
	cin >> n >> m;
	vector<int> b(n + 1, 1);
	inc(i, m) {
		int a;
		cin >> a;
		b[a] = 0;
	}
	
	incII(i, 0, n) { dp[i] = b[i] * (i < 2 ? 1 : dp[i - 1] + dp[i - 2]) % MOD; }
	
	cout << dp[n] << endl;
	
	return 0;
}