#include<iostream>
#include<fstream>
#include<cmath>

using namespace std;

int main(){
    ifstream cin("tmp.txt");
    int m,n;
    cin>>m>>n;
    while(m!=0||n!=0){    
        int i;
        int s[200];
        int j=0;
        for(i=0;i<n;i++){
            s[j++]=m-i;
        }    
        __int64 res=1;
        for(i=0;i<n;i++){          //从m到m-n+1里面找到可以被n-i整除的，然后除掉n-i
            for(j=0;j<n;j++){
                if(s[j]%(n-i)==0){
                    s[j]/=(n-i);
                    break;
                }
            }            
                    
        }
        for(i=0;i<n;i++){
            res*=s[i];
        }    
        cout<<m<<" things taken "<<n<<" at a time is "<<res<<" exactly."<<endl;
        //printf("%d things taken %d at a time is %I64d exactly.\n",dm,dn,res); 
        cin>>m>>n; 
    }
    system("pause");
    return 0;
}