docs: @Trivial
      private static String encode(String s) {
          if (s == null) {
              return null;
          }
  
          // Throw an IllegalArgumentException if "%21" is already present in the String
          if (s.contains("%21")) {
              throw new IllegalArgumentException("WSJPAURLUtils.encode() cannot encode Strings containing \"%21\".");
          }
          return s.replace("!", "%21");
      }