Number Puzzle

---

Time Limit: 2 Seconds      

Memory Limit: 65536 KB

---

Given a list of integers (A₁, A₂, ..., A_n), and a positive integer M, please find the number of positive integers that are not greater than M and dividable by any integer from the given list.

Input

The input contains several test cases.

For each test case, there are two lines. The first line contains N (1 <= N <= 10) and M (1 <= M <= 200000000), and the second line contains A₁, A₂, ..., A_n(1 <= A_i <= 10, for i = 1, 2, ..., N).

Output

For each test case in the input, output the result in a single line.

Sample Input

3 2

2 3 7

3 6

2 3 7

Sample Output

1

4

---

Author: 

MAO, Yiqiang

Source: 

Zhejiang University Local Contest 2007

1. 设 n, k 都是正整数，S={1,2,···,n}, 则 S 中能被 k 整除的正整数的个数为 [n/k].
2. 设 A_i (i=1,2,···,n) 为有限集，则
3. lcm(x,y)=xy/gcd(x,y)
4. lcm(x₁,x₂,···,x_n)=lcm(lcm(x₁,x₂,···,x_n-1),x_n)

#include <iostream>

#include <cstdio>

#include <cstring>

using namespace std;

int a[11],n,m,ans;

int gcd(int a,int b)

{

    return b==0?a:gcd(b,a%b);

}

int lcm(int a,int b)

{

    return a/gcd(a,b)*b;

}

void dfs(int x,int k,int t)

{

    if(x>0)  ans+=m/t*k;

    for(int i=x;i<n;i++)

        dfs(i+1,-k,lcm(a

,t));

}

int main()

{

while(cin>>n>>m)

{

  for(int i=0;i<n;i++)

      cin>>a

;

  ans=0;

  dfs(0,-1,1);

  cout<<ans<<endl;

}

    return 0;

}