BJFU fudq的等式

/*
 BJFU fudq的等式
 http://101.200.220.237/contest/19/problem/118/
 数论 勒让德定理 二分答案
 */
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
//#define test
using namespace std;
const int Nmax=1000005;
const long long INF=100000000000005LL;
long long p[Nmax];
long long times[Nmax];
int cnt;
long long n,x,y;

long long Legendre(long long n,long long p,long long y)
{
    long long ans=0LL;
    while(n>0)
    {
        ans+=n/p;
        if(ans>y)
            return y+1LL;
        n/=p;
    }
    return ans;
}

int is(long long n)
{
    for(int i=1;i<=cnt;i++)
    {
        if(Legendre(n,p[i],times[i]*y)<times[i]*y)
            return 0;
    }
    return 1;
}
long long get()
{
    long long n=0LL;
    long long l=1LL,r=INF;
    while(l<r)
    {
        long long mid=(l+r)>>1;
        long long m=is(mid);
        //printf("l:%lld r:%lld mid:%lld x:%lld y:%lld num:%lld
",l,r,mid,x,y,m);
        if(!m)
            l=mid+1;
        else 
            r=mid-1;
    }
    if(n==0LL && !is(r))
        n=r+1;
    else if(n==0LL && is(r))
        n=r;
    return n;

}
//long long get()
//{
    //long long m=-1LL;
    //for(int i=1;i<=cnt;i++)
    //{
       //m=max(m,get(p[i],times[i]*y)); 
    //}
    //return m;
//}
int main()
{
    #ifdef test
    #endif
    //freopen("test.in","r",stdin);
    while(scanf("%lld%lld",&x,&y)==2)
    {
        cnt=0;
        for(int i=2;i*i<=x;i++)
        {
            if(x%i==0)
            {
                p[++cnt]=i;
                times[cnt]=0;
                while(x%i==0)
                {
                    x/=i;
                    times[cnt]++;
                }
            }
        }
        if(x!=1)
        {
            p[++cnt]=x;
            times[cnt]=1;
        }

        if(cnt==0)
        {
            //printf("error
");
            printf("0
");
            continue;
        }
        printf("%lld
",get());
    }
    return 0;
}