Ошибка при запуске gulp.watch «SyntaxError: Identifier 'watch' has already been declared» как исправить?

const { series, parallel, src, dest } = require("gulp");
const browserSync = require('browser-sync').create();
const phpMinify = require('@cedx/gulp-php-minify');
const autoprefixer = require('gulp-autoprefixer');
const uglify = require('gulp-uglify');
const less = require('gulp-less');
const minifyCSS = require('gulp-csso');
const concat = require('gulp-concat');
const del = require('del');
const dir = {
    "src": {
        php: './app/*.php',
        html: './',
        css: './app/src/less/**/*.less',
        js: './app/src/js/**/*.js',
    },

    "dest": {
        build: './build/*',
        php: './build/',
        html: './',
        css: './build/css/',
        js: './build/js/',
    }
}

function php() {
    return src(dir.src.php, { read: false })
        .pipe(phpMinify({ mode: 'fast' }))
        .pipe(dest(dir.dest.php))
}

function css() {
    return src(dir.src.css)
        .pipe(less())
        .pipe(autoprefixer())
        .pipe(minifyCSS())
        .pipe(dest(dir.dest.css))
        .pipe(browserSync.stream())
}

function js() {
    return src(dir.src.js, { sourcemaps: true })
        .pipe(concat('app.min.js'))
        .pipe(uglify())
        .pipe(dest(dir.dest.js, { sourcemaps: true }))
        .pipe(browserSync.stream())
}

function clean() {
    return del(dir.dest.build)
}

function watch() {
    browserSync.init({
        server: {
            baseDir: './app'
        }
    })
        .pipe(watch(dir.src.css, css))
        .pipe(watch(dir.src.js, js))
        .pipe(watch(dir.src.php).on('change', browserSync.reload));
}

exports.js = js;
exports.css = css;
exports.php = php;
exports.clean = clean;
exports.watch = watch;
exports.build = series(this.clean, parallel(this.css, this.js))
exports.default = series(this.build, this.watch);