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 <vector>
#include <stack>
#include <functional>
#include <numeric>
#include <complex>
#include <queue>
#include <map>
#include <unordered_map>
#include <list>
#include <set>
#include <unordered_set>
#include <assert.h>
#include <cstring>
#include <sstream>
#include <string>
#include <algorithm>
#include <iostream>
 
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define X first
#define Y second
#define pb(x) push_back(x)
#define FOR(k, n) for (int k = 0; k < n; k++)
#define FORN(k, s, n) for (int k = s; k < n; k++)
#define debug(x) printf("x = %d\n", x);
#define debug2(x, y) printf("x = %d, y = %d\n", x, y);
#define debug3(x, y, z) printf("x = %d, y = %d, z = %d\n", x, y, z);
#define debug4(x, y, z, t) printf("x = %d, y = %d, z = %d, t = %d\n", x, y, z, t);
 
#pragma warning(disable : 6031)
#pragma warning(disable : 4101)
 
using namespace std;
 
typedef long long ll;
typedef pair<int, int> pii;
typedef long double ld;
typedef complex<double> cd;
 
const int INF = 1e9;
const long long LINF = 1e18;
const double PI = acos(-1);
 
// Solution

const int MAXN = 1000005;
const ll MOD = 1000000007ll;

ll n, m, x, A[MAXN], DP[MAXN];

int main() {
    scanf("%lld%lld", &n, &m);

    for (int i = 0; i < m; i++) {
        scanf("%lld", &x);
        A[x] = 1;
    }

    if (n > 1 && !A[n - 1]) {
        DP[n - 1] = 1;
    }

    if (n > 2 && !A[n - 2]) {
        DP[n - 2] = 1 + DP[n - 1];
    }

    for (int i = n - 3; i >= 0; i--) {
        if (!A[i]) {
            DP[i] = (DP[i + 1] + DP[i + 2]) % MOD;
        }
    }

    printf("%lld\n", DP[0]);

    return 0;
}