Problem:
Score : 100 points

Problem Statement

Takahashi has K 500-yen coins. (Yen is the currency of Japan.)
If these coins add up to X yen or more, print Yes; otherwise, print No.

Constraints

1 \leq K \leq 100

1 \leq X \leq 10^5

Input

Input is given from Standard Input in the following format:

K X

Output

If the coins add up to X yen or more, print Yes; otherwise, print No.

Sample Input 1

2 900

Sample Output 1

Yes

Two 500-yen coins add up to 1000 yen, which is not less than X = 900 yen.

Sample Input 2

1 501

Sample Output 2

No

One 500-yen coin is worth 500 yen, which is less than X = 501 yen.

Sample Input 3

4 2000

Sample Output 3

Yes

Four 500-yen coins add up to 2000 yen, which is not less than X = 2000 yen.

Solution in PyPy3 (2.4.0) (judge verdict: Runtime Error)

import fractions # version 3.4
from collections import defaultdict
from functools import reduce

def gcd(*numbers):
    return reduce(fractions.gcd, numbers)

def lcm(*numbers):
    ans = numbers[0]
    for a in numbers:
        ans = ans * a // fractions.gcd(ans, a)
    # return reduce(lambda x,y: (x * y) // fractions.gcd(x, y), numbers, 1)

    return ans


def s_in():
    return input()

def n_in():
    return int(input())

def l_in():
    return list(map(int, input().split()))

class Interval():
    def __init__(self, li):
        self.li = li
        self.n = len(li)
        self.sum_li = [li[0]]
        for i in range(1, self.n):
            self.sum_li.append(self.sum_li[i-1] + li[i])

    def sum(self, a, b=None):
        if b is None:
            return self.sum(0, a)

        res = self.sum_li[min(self.n-1, b-1)]
        if a > 0:
            res -= self.sum_li[a-1]
        return res

N, M = l_in()
A = l_in()

B = [a//2 for a in A]
L = lcm(*B)

if any((L//b)%2 == 0 for b in B):
    print(0)
else:
    print((M//L + 1)//2)