Have any code or way to loading generateblocks with condition?

Hello! When I check with Ctrl + U I see the post of used this code:
<style id="generateblocks-css">.gb-container .wp-block-image img{vertical-align:middle;}.gb-container .gb-shape{position:absolute;overflow:hidden;pointer-events:none;line-height:0;}.gb-container .gb-shape svg{fill:currentColor;}.gb-container.gb-container-845f5aca > .gb-inside-container{padding:30px 0;max-width:1200px;margin-left:auto;margin-right:auto;}.gb-container.gb-container-1865c68e > .gb-inside-container{padding:40px 0;max-width:1200px;margin-left:auto;margin-right:auto;}.gb-icon{display:inline-flex;line-height:0;}.gb-icon svg{height:1em;width:1em;fill:currentColor;}.gb-highlight{background:none;color:unset;}@media (max-width: 767px) {.gb-grid-wrapper > .gb-grid-column{width: 100%;}}</style>
But I think this code not necessary because this is post, not custom page and I not use and block of plugin Generate Block.
So have any code or way to loading generateblocks with condition? I don't want CSS of Generate Block load in post!
I was add code like this but it not useful:

add_action('wp_enqueue_scripts', 'remove_css_generateblocks_post', 99);
function remove_css_generateblocks_post() {
	if(function_exists('is_generateblock')) {
                if(is_single()){
                wp_dequeue_style( 'generateblocks' );
                }
	}
}