int mymod(int m, int n){
       //this function will return m%n;
       int n2, n3, tmp, again, kk;
       n2=0; //while(n2)--n2;
       n3=0; //while(n3)--n3;
       tmp=0;  //while(tmp)--tmp;
       again=0; //while(again)--again;  // again = 0;
       ++again;  // again = 1
       while(again) {
          while(n){--n; ++n2; ++tmp;}  // save n in tmp
          while(n2){
             --n2;
             --m;
             kk = invert(m);
             while(kk) {   // kk ==1 means m = 0
                --kk; // kk=0;
                --again;  // again = 0;
             }
          }//while n2
          while(tmp){ --tmp; ++n; } // recover n
       }// again
       // if m !=0 then let m = m + n
       kk = invert(m);
       while(kk) {  // kk==1 means m = 0;
           --kk; // kk = 0;
           n=0; //while(n) --n; // n = 0
       }
       while(n) { --n; ++m; } // 因 m 減過頭要加回去
       return m;  // this is the answer of  m % n
   } //mymod(