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: Runtime Error)

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <functional>
#include <map>
#include <iomanip>
#include <math.h> 
#include <stack>
#include <queue>
#include <bitset>
#include <cstdlib>
#include <tuple>
#include <cctype>
#include <ctype.h>
#include <set>
#include <sstream>
#include <time.h>
using namespace std;
#define int long long
#define rep(i,s,n) for(int i = s;i<n;i++)
#define repe(i,s,n) for(int i = s;i<=n;i++)
#define rrep(i,s,n) for(int i = (n)-1;i>=(s);i--)
#define all(v) (v).begin(),(v).end()
#define pb push_back
#define fi first
#define se second
#define chmin(a,b) a=min((a),(b))
#define chmax(a,b) a=max((a),(b))
#define l1 list[index]
#define l2 list[index - 1]
#define l3 list[index + 1]
#define iif(i,j) ((i<0 && j<0) || (i>0 && j>0)) ? true : false 
typedef long long ll;
typedef pair<int, int>pint;
typedef vector<int>vint;
typedef vector<pint>vpint;
typedef pair<pint, int> P1;
typedef pair<int, pint> P2;
typedef pair<pint, pint> PP;
static const ll maxLL = (ll)1 << 62;
const ll MOD = 1000000007;
const ll INF = 1e18;
const double PI = 3.14159265359;
int ca[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };



signed main() {
	int n, m;
	vector<int>list(105, -1);
	bool flag[100005];
	int dp[100005];

	memset(dp, 0, sizeof(dp));
	memset(flag, false, sizeof(flag));

	cin >> n >> m;

	for (int i = 0; i < m; i++) {
		cin >> list[i];
		flag[list[i]] = true;
	}

	dp[0] = 1;

	if (flag[1] == false) {
		dp[1] = 1;
	}
	else {
		dp[1] = 0;
	}

	for (int i = 2; i <= n + 1; i++) {
		//if( flag[i] == false) dp[i] = dp[i - 1] + 1;
		if (flag[i] == false) dp[i] += dp[i - 1];

		dp[i] %= MOD;
		//if (flag[i] == false) dp[i] = dp[i - 1] * dp[i];
		//if( flag[i] == false) dp[i] = dp[i - 2] + 1;
		if (flag[i] == false ) dp[i] += dp[i - 2];

		dp[i] %= MOD;
	}


	for (int i = 0; i <= n + 1; i++) {
		//cout << "i->" << i << " " << dp[i] << endl;
	}

	cout << dp[n] << endl;




	getchar();
	getchar();
	return 0;
}