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: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int dp(int n){
    if(n == 1){
        return 1;
    }
    else if(n == 2){
        return 1;
    }
    else if(n == 3){
        return 2;
    }
    else if(n >= 4){
        return dp(n-1)+dp(n-2);
    }
    else{
        printf("dp() seems has the problem\n");
        return 0;
    }
}

int main(){
    int i,j,N,M;
    int a[100000];
    unsigned long sum=0;
    int current=0;
    int numOfSteps=0;
    int flg=0;

    scanf("%d %d", &N, &M);
    for(i=0;i<M;i++){
        scanf("%d", &a[i]);
    }

    for(i=1;i<M;i++){
        // finish if broken steps are continuous
        if(a[i]-a[i-1] == 1) {
            printf("0\n");
            return 0;
        }
    }

    current=0;
    sum=1;
    for(i=0;i<=N;i++){
        for(j=current;j<M;j++){
            if(i==a[j]){
                sum *= dp(numOfSteps);
                sum = sum % 1000000007;
                flg=1;
                current++;
            }
        }
        if(flg==1){
            flg=0;
            numOfSteps=0;
        }
        else{
            numOfSteps++;
        }
    }
    if(numOfSteps != 0){
        sum *= dp(numOfSteps);
        sum = sum % 1000000007;
    }

    printf("%lu\n", sum);
    return 0;
}