.breadcrumb {
	/*centering*/
	display: inline-block;
	overflow: hidden;
	border-radius: 5px;
	/*Lets add the numbers for each link using CSS counters. flag is the name of the counter. to be defined using counter-reset in the parent element of the links*/
	counter-reset: flag;
	transform: scale(1.3);
	margin-top: 1em;
	margin-bottom: 0.5em;
	margin-left: 33%;
	/*width: 15%;*/
	position: relative;
	top: 5px;
}
div.breadcrumb-single-element {
    /*display: block;
    margin-right: auto;
    margin-left: auto;*/
}
.breadcrumb > a {
    text-decoration: none;
    outline: none;
    display: block;
    float: left;
    font-size: 12px;
    line-height: 36px;
    color: white;
    /*need more margin on the left of links to accomodate the numbers*/
    padding: 0 20px 0 30px;
    /*background: #666;
	background: linear-gradient(#666, #333);*/
    position: relative;
    border: 1px solid gray;
}
/*since the first link does not have a triangle before it we can reduce the left padding to make it look consistent with other links*/
.breadcrumb > a:first-child {
	padding-left: 20px;
	border-radius: 5px 0 0 5px; /*to match with the parent's radius*/
}
.breadcrumb > a:first-child:before {
	left: 14px;
}
.breadcrumb > a:last-child {
	border-radius: 0 5px 5px 0; /*this was to prevent glitches on hover*/
}

/*.breadcrumb > a:before {
	content: counter(flag);
	counter-increment: flag;
	border-radius: 100%;
	width: 20px;
	height: 20px;
	line-height: 20px;
	margin: 8px 0;
	position: absolute;
	top: 0;
	left: 30px;
	font-weight: bold;    
    padding-left: 0.55em;
}*/

.breadcrumb > a:after {
	content: '';
	position: absolute;
	top: 0; 
	right: -18px; /*half of square's length*/
	/*same dimension as the line-height of .breadcrumb > a */
	width: 36px; 
	height: 36px;
	/*as you see the rotated square takes a larger height. which makes it tough to position it properly. So we are going to scale it down so that the diagonals become equal to the line-height of the link. We scale it to 70.7% because if square's: 
	length = 1; diagonal = (1^2 + 1^2)^0.5 = 1.414 (pythagoras theorem)
	if diagonal required = 1; length = 1/1.414 = 0.707*/
	transform: scale(0.707) rotate(45deg);
	/*we need to prevent the arrows from getting buried under the next link*/
	z-index: 1;
	/*background same as links but the gradient will be rotated to compensate with the transform applied*/
	/*background: #666;
	background: linear-gradient(135deg, #666, #333);*/
	/*stylish arrow design using box shadow*/
	box-shadow: 
		2px -2px 0 2px rgba(0, 0, 0, 0.4), 
		3px -3px 0 2px rgba(255, 255, 255, 0.1);
	/*
		5px - for rounded arrows and 
		50px - to prevent hover glitches on the border created using shadows*/
	border-radius: 0 5px 0 50px;
}
/*we dont need an arrow after the last link*/
.breadcrumb > a:last-child:after {
	content: none;
}