Ошибка org.hibernate.PropertyAccessException

package sed.backend;
 
import javax.persistence.*;
 
@Entity
@Table(name = "Person")
public class Record {
 
    public Record(){
 
    }
 
    @Id
    @GeneratedValue
    @Column(name="id")
    private Long id;
    public Long getId(){return id;}
    public void setId(Long id){this.id = id;}
 
    @Column(name = "uname")
    private String uname;
    public String getUname(){return uname;}
    public void setUname(String uname){this.uname = uname;}
 
}
 
package sed.backend;
 
import javax.ws.rs.ApplicationPath;
 
@ApplicationPath("/api")
public class MainPoint extends javax.ws.rs.core.Application{
 
}
 
package sed.backend;
 
import org.hibernate.Session;
import org.hibernate.SessionFactory;
import org.hibernate.Transaction;
 
import javax.ejb.Stateful;
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
import javax.ws.rs.GET;
import javax.ws.rs.Path;
import javax.ws.rs.Produces;
import javax.ws.rs.core.MediaType;
import java.util.List;
 
@Stateful
@Path("/controller")
public class Controller {
 
    @PersistenceContext(unitName = "tunit")
    private EntityManager em;
 
    @Path("sayHello")
    @GET
    @Produces(MediaType.TEXT_PLAIN)
    public String sayHello(){
 
        Session session = em.unwrap(Session.class);
 
        SessionFactory sessionFactory = session.getSessionFactory();
        Session currentSession = sessionFactory.openSession();
        Transaction currentTransaction = session.beginTransaction();
 
        Record record = new Record();
        record.setUname("uname1");
 
        em.persist(record);
 
        currentTransaction.commit();
        currentSession.close();
 
        List<Record> records = (List<Record>) currentSession.createQuery("from Record").list();
 
        return Integer.toString(records.size());
    }
 
}