Проблеммы с gulp

Выдает ошибку Еype Error: cannot read property 'html' введите сюда описание изображенияof undefined

let project_folder="dist";
let sourse_folder="#src";

let path = {
  buid: {
    html: project_folder + "/",
    css: project_folder + "/css/",
    js: project_folder + "/js/",
    img: project_folder + "/img/",
    fonts: project_folder + "/fonts/",
  },
  src: {
    html: sourse_folder + "/*.html",
    css: sourse_folder + "/scss/style.scss",
    js: sourse_folder + "/js/script.js",
    img: sourse_folder + "/img/**/*.{jpg,png,svg,gif,ico,webp}",
    fonts: sourse_folder + "/fonts/*.ttf",
  },
  watch: {
    html: sourse_folder + "/**/*.html",
    css: sourse_folder + "/scss/**/*.scss",
    js: sourse_folder + "/js/**/*.js",
    img: sourse_folder + "/img/**/*.{jpg,png,svg,gif,ico,webp}",
  },
  clean: "./" + project_folder + "/"
}

let { src, dest } = require('gulp'),
  gulp = require('gulp'),
  browsersync = require('browser-sync').create();

  function browserSync(params) {
    browsersync.init({
      server: {
        baseDir: "./" + project_folder + "/"
      },
      port: 3000,
      notify: false
    })
  }

  function html() {
    return src(path.src.html)
      .pipe(dest(path.build.html))
      .pipe(browsersync.stream())
  }

let build = gulp.series(html);
let watch = gulp.parallel(build,browserSync);

exports.html = html;
exports.build = build;
exports.watch = watch;
exports.default = watch;


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