Как вернуть исходный src img

$(document).ready(function() { 
  var items = $('.sort-item').find('img'); 
 
  items.click(function() { 
    var clicked = $(this); 
 
    items.each(function() { 
      var el = $(this), 
        original = el.data('original'); 
 
      if (el.attr('src') !== clicked.attr('src')) { 
        el.attr('src', original); 
      } 
    }); 
 
    var el = $(this), 
      original = el.data('original'), 
      alt = el.data('alt'); 
 
    if (el.attr('src') === alt) { 
      el.attr('src', original); 
    } else { 
      el.attr('src', alt); 
    } 
  }); 
});

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 
<div class="sort"> 
  <a href="#" class="sort-item" title="Безалкогольная продукция"> 
    <img src="http://lorempixel.com/output/food-q-g-100-100-7.jpg" data-original="http://lorempixel.com/output/food-q-g-100-100-7.jpg" data-alt="http://lorempixel.com/output/food-q-c-100-100-8.jpg"> 
  </a> 
  <a href="#" class="sort-item" title="Безалкогольная продукция"> 
    <img src="http://lorempixel.com/output/food-q-g-100-100-2.jpg" data-original="http://lorempixel.com/output/food-q-g-100-100-2.jpg" data-alt="http://lorempixel.com/output/food-q-c-100-100-1.jpg"> 
  </a> 
  <a href="#" class="sort-item" title="Безалкогольная продукция"> 
    <img src="http://lorempixel.com/output/food-q-g-100-100-4.jpg" data-original="http://lorempixel.com/output/food-q-g-100-100-4.jpg" data-alt="http://lorempixel.com/output/food-q-c-100-100-6.jpg"> 
  </a> 
</div>