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 (Clang 3.8.0) (judge verdict: Accepted)

#define D(x) cout << #x " = " << x << endl
#define DD(x, y) cout << "(" #x ", " #y ") = (" << x << ", " << y << ")" << endl
#define DDD(x, y, z) cout << "(" #x ", " #y ", " #z ") = (" << x << ", " << y << ", " << z << ")" << endl
#define DV(v) for (auto __macro_vi : v){ cout << __macro_vi << " ";} cout << endl;
#define DA(a, n) for (int __macro_i = 0; __macro_i < n; ++__macro_i) { cout << a[__macro_i] << " "; } cout << endl;
#define REP(i, n) for (int i = 0; i < n; ++i)
#define ALL(v) (v).begin(), (v).end()
#include<iostream>
#include<vector>
#include<cassert>
#include<cmath>

using namespace std;

using ll = long long;
const int MOD = 1000000007;
ll dp[100002];
bool broken[100002];
int n, m;
int main()
{
    cin >> n >> m;
    fill(broken, broken + n + 1, false);
    fill(dp, dp + n + 1, 0);
    dp[0] = 1;
    REP(i, m) {
        int ai;
        cin >> ai;
        broken[ai] = true;
    }
    for (int i = 1; i <= n; ++i) {
        dp[i] = 0;
        if (broken[i]) continue;
        if (i - 2 >= 0) dp[i] = (dp[i] + dp[i - 2]) % MOD;
        dp[i] = (dp[i] + dp[i - 1]) % MOD;
    }
    cout << dp[n] << endl;
    // DA(dp, 10);
}