How to show the previous page gridview images in flipview control in another page in windows 8? -

i displaying items in grid view .i want show grid view items in page using flip view control.how dynamically display selected item position in second page ? please tell me how achieve this?

edit: in first page : grid view item click event wrote code this:

private void photogrid_itemclick(object sender, itemclickeventargs e)         {             var itemid = ((flipimage)e.clickeditem);             flipimage s = new flipimage() { imageurl = itemid.imageurl, title = itemid.title };             this.frame.navigate(typeof(flippage), s);         } 

in second page:

protected override void onnavigatedto(navigationeventargs e)         {              flipimage s = (flipimage)e.parameter;            string url =  s.imageurl;            flipviewcontrol.items.add(url);           } 

i want display previous page selected item in second page , click on next in flipview need show after selected item data.please tell me how write code. data binding flipview :

xdocument xdoc = xdocument.load("xmlfile1.xml");             ienumerable<flipimage> images = img in xdoc.descendants("image") select new flipimage(img.element("imagetitle").value, img.element("imageurl").value);             flipviewcontrol.datacontext = images; 

design of flipview:

<grid background="{staticresource applicationpagebackgroundthemebrush}">         <flipview horizontalalignment="left" verticalalignment="top" x:name="flipviewcontrol" itemssource="{binding}">             <flipview.itemtemplate>                 <datatemplate>                      <image horizontalalignment="left"  source="{binding imageurl}" height="762" verticalalignment="top" width="1360" x:name="imagecontrol"/>                 </datatemplate>             </flipview.itemtemplate>          </flipview>      </grid> 

please tell me how show previous page selected item value , when click on next in filpview need show after selected items data of page vice versa!!!