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

n, m = map(int, input().split())
broken = set([int(input()) for _ in range(m)]) # ここが'list'だとPypyでも通らない

def ptn_num(x): # 'x'stepまでのパターン数
    dp = [0] * (x+1) # DP枠_including step'0'
    if 1 in broken: # in case, step'1' broken 
        dp[:2] = 1, 0
    else:
        dp[:2] = 1, 1 # Usually

    for i in range(2, x+1):
        if i not in broken: # 階段が壊れてない場合...
            dp[i] = (dp[i-1] + dp[i-2]) % (10**9+7) # -1段目と-2段目の合計
    print(dp[-1])

ptn_num(n)