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

#include <math.h>

#define mod 1000000007

long long int Fibo(int n) {
	return round((pow((1 + sqrt(5)) / 2, n)
		- pow((1 - sqrt(5)) / 2, n)) / sqrt(5));
}


int main(){

	int i, j, k, N, M, *a;	
    long long int sum = 1, result;

	scanf("%d %d", &N, &M);

	if (M == 0){//危険マスがない場合はスタートからゴールまですぐ計算
		printf("%lld\n", Fibo(N) % mod);
		return 0;
	}

	a = (int *)malloc(M * sizeof(int));

	for (i = 0; i < M; i++)
		scanf("%d", &a[i]);
	
	if (M != 1){ // 連続した危険マスが隣接していたら０を返す
		for (i = 0; i < M; i++){
			if (a[i + 1] - a[i] == 1){
				printf("0\n");
				return 0;
			}
		}
	}

	sum *= Fibo(a[0]);//スタートから最初のブロック(直前)までの総数
	sum %= mod;

	for (i = 0; i < M-1; i++){//最初のブロックから最後のブロックまでの総数計算
		sum *= Fibo(a[i + 1] - a[i]-1);
		sum %=mod;
	}

	sum *= Fibo(N - a[M - 1]);//最後のブロックからゴールまでの総数

	result = sum % mod;
	printf("%lld\n", result);

  return 0;
}