Now you get a number N, and a M-integers set, you should find out how many integers which are small than N, that they can divided exactly by any integers in the set. For example, N=12, and M-integer set is {2,3}, so there is another set {2,3,4,6,8,9,10}, all the integers of the set can be divided exactly by 2 or 3. As a result, you just output the number 7.

Input

  There are a lot of cases. For each case, the first line contains two integers N and M. The follow line contains the M integers, and all of them are different from each other. 0<N<2^31,0<M<=10, and the M integer are non-negative and won’t exceed 20.

Output

  For each case, output the number.

Sample Input

12 2
2 3

Sample Output

7

題解：發現數據並不大最多10個，狀壓一下，容斥即可

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define lowbit(x) (x&(-x))
#define INF 0x3f3f3f3f
const int N=12;
typedef long long ll;
ll n,m,a[N];
int main()
{

    while(scanf("%lld%lld",&n,&m)!=EOF)
    {
        int len=0;
        for(ll i=1;i<=m;i++)
        {
            ll x;
            scanf("%lld",&x);
            if(x) a[++len]=x;
        }
        m=len;
        ll mm=(1<<m)-1;
        ll ans=0;
        n--;
        for(ll i=1;i<=mm;i++)
        {
            ll tmp=i;
            ll cnt=0;
            ll res=1;
            while(tmp)
            {
                if(tmp&1) cnt++;
                tmp/=2;
            }
            for(int j=1;j<=m;j++)
            {
                if(i&(1<<(j-1)))
                {
                    res=res*a[j]/__gcd(res,a[j]);
                }
            }
            if(cnt&1) ans+=n/res;
            else ans-=n/res;
        }
        printf("%lld\n",ans);
    }
	return 0;
}