Like expression in query

I have to create a filter for table based on the string from text field. From my controller I create the query string like this 'SELECT e FROM seabis$TaskQueueItem e WHERE e.title LIKE :(?!)custom$title ’ and set it to collection DS linked with the table. The custom$title is equal to string from text field. So I made all steps from here http://docs.cuba-platform.com/cuba/5.6/manual/en/webhelp/datasource_query_case_insensitive.html but got the exception:
‘ArgumentException: Encountered “e . title LIKE : (”…’

    at org.apache.openjpa.kernel.jpql.JPQL.generateParseException(JPQL.java:13180)
    at org.apache.openjpa.kernel.jpql.JPQL.jj_consume_token(JPQL.java:13054)
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_primary(JPQL.java:1980)
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_factor(JPQL.java:1958)
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_term(JPQL.java:1807)
    at org.apache.openjpa.kernel.jpql.JPQL.conditional_expression(JPQL.java:1769)
    at org.apache.openjpa.kernel.jpql.JPQL.where_clause(JPQL.java:1587)
    at org.apache.openjpa.kernel.jpql.JPQL.select_statement(JPQL.java:91)
    at org.apache.openjpa.kernel.jpql.JPQL.parseQuery(JPQL.java:63)
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder$ParsedJPQL.parse(JPQLExpressionBuilder.java:2412)
    at org.apache.openjpa.kernel.jpql.JPQLExpressionBuilder$ParsedJPQL.<init>(JPQLExpressionBuilder.java:2399)
    at org.apache.openjpa.kernel.jpql.JPQLParser.parse(JPQLParser.java:49)

Am I miss something?

You have mistyped the prefix: use “(?i)” instead of “(?!)”

Oh… i instead of !. Thanks!