<pre>
long a<code>10000, b, c</code>2800, d, e, f~[[2801]], g;

main()
<pre>
   {
    for (;b-c;) f~[[b++]]=a/5;

    for (;
	d<code>0,g</code>c*2;
	c-<code>14, printf("%.4d",e+d/a), e</code>d%a)

    for (b=c;
	 d+<code>f~[[b]]*a,f~[[b]]</code>d%--g,d/=g--,--b;
	 d*=b);
   }
</pre>
</pre>
