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 Python (3.4.3) (judge verdict: Accepted)

import sys
input = sys.stdin.readline


def binary_search(l, value):
    low = 0
    high = len(l) - 1
    while low <= high:
        mid = (low + high) // 2
        if l[mid] == value:
            return True
        elif l[mid] < value:
            low = mid + 1
        else:
            high = mid - 1
    return False


N, M = map(int, input().split())
A = [int(input()) for _ in range(M)]

A.append(10**6)
A.append(10**6)
B = [0] * (N + 1)
B[0] = 1
if binary_search(A, 1) is True:
    B[1] = 0
else:
    B[1] = 1
mod = 10**9 + 7

for i in range(2, N + 1):
    if i != 2 and binary_search(A, i - 1) is True and binary_search(
            A, i - 2) is True:
        print(0)
        exit()
    elif binary_search(A, i - 2) is True:
        B[i] = B[i - 1]
    elif binary_search(A, i - 1) is True:
        B[i] = B[i - 2]
    else:
        B[i] = B[i - 1] + B[i - 2]
    B[i] %= mod
print(B[N] % mod)