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>
int combination(int n, int r)
{
	if (r == 0 || r == n)
		return (1);
	else if (r == 1)
		return (n);
	return (combination(n - 1, r - 1) + combination(n - 1, r));
}
int main() {
	int n,m,i,j,x,t,a[100000];
	long long int ans=1;
	scanf("%d", &n);
	scanf("%d", &m);
	a[0] = -1;
	for (i = 1; i <= m; i++) {
		t = 0;
		scanf("%d", &a[i]);
		x=a[m] - a[m - 1] - 1;
		for (j = 0; x >= j * 2;j++) t += combination(x - j, j);
		ans *= t;
	}
	t = 0;
	x = n - a[m]-1;
	for (j = 0; x >= j * 2; j++) t += combination(x - j, j);
	printf("%lld", ans*t%1000000007);
}