if n is odd then every set will contain 2 adjacent element except the last set which will have last three elements
for ( int i = 0 ; i < n / 2 - 1 ; i ++ ) { firstadj = 2 * i + 1 ; secadj = 2 * i + 2 ; cout << " ( " << firstadj << " , ▁ " << secadj << " ) STRNEWLINE " ; }