如何为只有一个项目和一个列表的XML元素实现POJO？

import javax.xml.bind.annotation.XmlElement;
import javax.xml.bind.annotation.XmlRootElement;

@XmlRootElement(name = "RootElement")
public class MyPOJO {
    private String project;
    private List<String> items;

    public String getProject() {
        return project;
    }

    @XmlElement(name = "Project")
    public void setProject(String project) {
        this.project = project;
    }

    public List<String> getItems() {
        return items;
    }

    @XmlElement(name = "Item")
    public void setItems(List<String> items) {
        this.items = items;
    }
}

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Unmarshaller;
import java.io.File;

public class XMLToObject {
    public static void main(String[] args) {
        try {
            File file = new File("path/to/xml/file.xml");
            JAXBContext jaxbContext = JAXBContext.newInstance(MyPOJO.class);
            Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
            MyPOJO myPOJO = (MyPOJO) jaxbUnmarshaller.unmarshal(file);
            System.out.println(myPOJO.getProject());
            System.out.println(myPOJO.getItems());
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}

import javax.xml.bind.JAXBContext;
import javax.xml.bind.JAXBException;
import javax.xml.bind.Marshaller;
import java.io.File;
import java.util.Arrays;

public class ObjectToXML {
    public static void main(String[] args) {
        try {
            MyPOJO myPOJO = new MyPOJO();
            myPOJO.setProject("Project Name");
            myPOJO.setItems(Arrays.asList("Item 1", "Item 2", "Item 3"));

            File file = new File("path/to/save/xml/file.xml");
            JAXBContext jaxbContext = JAXBContext.newInstance(MyPOJO.class);
            Marshaller jaxbMarshaller = jaxbContext.createMarshaller();
            jaxbMarshaller.setProperty(Marshaller.JAXB_FORMATTED_OUTPUT, true);
            jaxbMarshaller.marshal(myPOJO, file);
        } catch (JAXBException e) {
            e.printStackTrace();
        }
    }
}