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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>

typedef long long int ll;

ll comb(ll n, ll k){
    ll num = 1;
    ll i;
    for(i = 1; i <= k; i++){
        num = num*(n-i+1)/i;
    }
    return num;
}



int main(void){
    int M, N;
    ll a[100000];
    int i = 0;
    ll ans = 1;
    ll cnt = 0;
    ll div=1000000007;


    scanf("%d %d", &N, &M);
    for(i = 0; i < M; i++){
        scanf("%d", &a[i]);
        if(i != 0 && a[i-1]+1 == a[i]){
            printf("0");
            return 0;
        }
    }
    a[M] = N;

    ll numtwo = (a[0]-1)/2;
    ll numone = (a[0]+1)%2;

    for(i = 0; i <= M; i++){
        cnt = 0;
        if(i == 0){
            while(numtwo != 0){
                cnt += comb(numone+numtwo, (numone >= numtwo ? numtwo : numone))%div;
                numtwo--;
                numone += 2;
            }
            ans = ans*(cnt+1)%div;
            continue;
        }else if((a[i]-1)-(a[i-1]+1) >= 2 && i!= M){
            numtwo = (a[i]-a[i-1]-2)/2;
            numone = (a[i]-a[i-1]-2)%2;
            while(numtwo != 0){
                cnt += comb(numone+numtwo, (numone >= numtwo ? numtwo : numone))%div;
                numtwo--;
                numone += 2;
            }
            ans = ans*(cnt+1)%div;
        }else if(i == M){
            numtwo = (a[M] - a[M-1] - 1)/2;
            numone = (a[M] - a[M-1] - 1)%2;
            while(numtwo != 0){
                cnt += comb(numone+numtwo, (numone >= numtwo ? numtwo : numone))%div;
                numtwo--;
                numone += 2;
            }
            ans = ans*(cnt+1)%div;
        }
    }

    printf("%lld", ans%div);
    return 0;
}