When you have multiple CSS entry points, how do you NOT rebuild all files if only one of them changes?

Question

JavaScript

The project consists of several parts that are independent of each other,
so it is logical to single out several entry points, in particular for css.
There is a task for building css.
How not to rebuild all files if only one of them changes?
And then rebuilding all the files greatly slows down the work.

const path.src.style = ['./assets/src/style/main.scss', './assets/src/style/another-point.scss'],
          path.build.css= './assets/build/css/';
// style build
function css_build () {
    return gulp.src(path.src.style)
        .pipe(plumber())
        .pipe(gulpif(devMode, sourcemaps.init()))
        .pipe(sass())
        .pipe(autoprefixer({
            overrideBrowserslist:  ['last 2 versions'],
            cascade: false
        }))
        .pipe(gulp.dest(path.build.css))
        .pipe(rename({ suffix: '.min' }))
        .pipe(cleanCSS({level: {1: {specialComments: 0}}}))
        .pipe(gulpif(devMode, sourcemaps.write('./'))) 
        .pipe(gulp.dest(path.build.css))
        .pipe(browserSync.reload({ stream: true }));
}
exports.css_build = css_build;