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: Accepted)

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

#define DIV 1000000007

int main(void){
	long long n, m;
	long long a[200000] = { 0 };
	long long ans = 1;
	long long f[200000] = { 0 };
	int flag = 0;

	scanf("%lld %lld", &n, &m);
	for (int i = 0; i < m; i++) {
		scanf("%lld", &a[i]);
		if (i && a[i] == a[i - 1] + 1) flag = 1;
	}
	if (flag)printf("0");
	else {
		f[0] = 1, f[1] = 1;
		for (int i = 2; i <= n; i++)
			f[i] = (f[i - 1] % DIV + f[i - 2] % DIV) % DIV;

		if (m == 0)printf("%lld", f[n]);
		else {
			for (int i = 0; i <= m; i++) {
				if (i == 0)ans = (ans * f[a[i] - 1]) % DIV;
				else if (i == m)ans = (ans * f[n - a[i - 1] - 1]) % DIV;
				else ans = (ans * f[a[i] - a[i - 1] - 2]) % DIV;
			}
			printf("%lld", ans);
		}
	}
	return 0;
}