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)

#!/usr/bin/env python
# -*- coding: utf-8 -*-

N, M = map(int, input().split())
# 0段目からN段目まである
N += 1
broken_stairs = [int(input()) for _ in range(M)]

# dp[i] = i段目に辿り着くまでの経路本数
dp = [1 for _ in range(N)]

# 初期化する
for b in broken_stairs:
    dp[b] = 0

# MOD用
MOD = 10 ** 9 + 7

for i in range(2, N):
    """TLEだったやつ
    # 1段上が範囲内で使用可能なら現在の段の経路数を足す
    if i + 1 < N and i + 1 not in broken_stairs:
        # MODを取って更新
        dp[i + 1] += dp[i] % MOD
    # 2段上が範囲内で使用可能なら現在の段の経路数を足す
    if i + 2 < N and i + 2 not in broken_stairs:
        # MODを取って更新
        dp[i + 2] += dp[i] % MOD
    """
    # i段目が使用可能なら1段前と2段前の経路数を足した値を入れる
    if dp[i]:
        dp[i] = (dp[i - 1] + dp[i - 2]) % MOD


print(dp[-1])