#include <iostream>
#include <fstream>
using namespace std;
ifstream fin ("mins.in");
ofstream fout ("mins.out");
long long c,d,k,cnt[200005],nrp;
int main()
{
    fin >>c>>d;
    for (int k=2;k<=200000;k++)
        {nrp-=(cnt[k]-1)*(c-1)/k*(d-1)/k;
         for (int m=2*k;m<=200000;m=m+k)
             {cnt[m]-=(cnt[m]-1);}}
    fout <<(c-1)*(d-1)-nrp;
    return 0;
}