如何获取XSSFSimpleShape对象的背景颜色？

import org.apache.poi.ss.usermodel.*;
import org.apache.poi.xssf.usermodel.*;

public class GetSimpleShapeBackgroundColor {
    public static void main(String[] args) throws Exception {
        // 加载Excel文件
        Workbook workbook = new XSSFWorkbook("example.xlsx");

        // 获取第一个工作表
        Sheet sheet = workbook.getSheetAt(0);

        // 获取绘图对象
        XSSFDrawing drawing = (XSSFDrawing) sheet.getDrawingPatriarch();

        // 获取所有图形对象
        for (XSSFShape shape : drawing.getShapes()) {
            // 判断是否为XSSFSimpleShape对象
            if (shape instanceof XSSFSimpleShape) {
                XSSFSimpleShape simpleShape = (XSSFSimpleShape) shape;
                // 获取背景颜色
                XSSFColor backgroundColor = simpleShape.getFillColor();
                if (backgroundColor != null) {
                    // 输出背景颜色的RGB值
                    byte[] rgb = backgroundColor.getRGB();
                    System.out.println("背景颜色的RGB值：" + rgb[0] + ", " + rgb[1] + ", " + rgb[2]);
                }
            }
        }

        // 关闭工作簿
        workbook.close();
    }
}