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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <stdio.h>
#include <vector>
#include <string.h>
#include <algorithm>
#include <map>
#include <string>
#include <deque>

#pragma warning(disable:4996)

int jun(int n) {
	int ret = 1;
	for (int i = n; i > 0; --i) {
		ret = (ret * i) % 1000000007;
	}
	return ret;
}

int calc(int n) {
	int cc = 0;
	int two_max = n / 2;
	for (int i = 0; i <= two_max; ++i) {
		int one = n - 2 * i;
		int junOne = jun(one);
		int junTwo = jun(i);
		int junAll = jun(one+i);
		int c = (junAll / (junOne * junTwo)) % 1000000007;
		cc = (cc + c) % 1000000007;
	}
	return cc;
}

int main(void) {
	int n, m;
	scanf("%d %d", &n, &m);
	std::vector<int> A;
	for (int i = 0; i < m; ++i) {
		int a;
		scanf("%d", &a);
		A.push_back(a);
	}
	A.push_back(n+1);
	int now = 0;
	int c = 1;
	for (int i = 0; i < m+1; ++i) {
		int l = A[i] - now - 1;
		if (l == 0) {
			printf("0");
			return 0;
		}
		c = (c * calc(l - 1))%1000000007;
		now = A[i] - 1;

	}

	printf("%d", c);
	return 0;
}