csu 1803(2016)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=3000;

/*
   n m <=1e9
   a*b=2016p  (0<a<=n  0<b<=m)
   的组合有多少个
   那么也就是a*b%2016=0%2016
   
   那么只要枚举2016*2016个就好了
*/

ll cnt1[maxn],cnt2[maxn];

void solve(int n,ll c[]){
    int a=n/2016,b=n%2016;
    for(int i=1;i<=b;i++)c[i]=a+1;
    for(int i=b+1;i<=2016;i++)c[i]=a;
}

int main(){
    ll n,m;
    
    while(cin>>n>>m){
    
        solve(n,cnt1);
        solve(m,cnt2);
        
        ll res=0;
        
        for(int i=1;i<=2016;i++)for(int j=1;j<=2016;j++)
            if((i*j%2016)==0){
                res+=cnt1[i]*cnt2[j];
            }
        cout<<res<<endl;
    }
    return 0;
}