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.

 

  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.

 

  For each case, output the number.

 

 

 

#include<bits/stdc++.h>

using namespace std;

const int N = 3e6+, M = 1e6+, mod = 1e9+,inf = 1e9+;

typedef long long ll;

ll ans;

int x,a[N],n,m;

ll gcd(ll a,ll b) {return b==?a:gcd(b,a%b);}

void dfs(int i,int num,ll tmp) {

     if(i>=m) {

        if(num==) ans=;

        else {

            if(num&) ans = (ans+n/tmp);

            else ans = ans - n/tmp;

        }

        return ;

     }

     dfs(i+,num,tmp);

     dfs(i+,num+,tmp*a[i]/gcd(tmp,a[i]));

}

int main() {

    while(scanf("%d%d",&n,&m)!=EOF) {

        int k = ;

        n--;

        for(int i=;i<=m;i++) {

            scanf("%d",&x);

            if(x) a[k++] = x;

        }

        m = k;

        ans = ;

        dfs(,,);

        printf("%I64d\n",ans);

    }

    return ;

}