xml.linq如何解析如下xml文件

xml文件结构

<?xml version="1.0" encoding="UTF-8" ?>

 <teacher LessonTitle="测试">

	<general title="你懂的" msg="不懂"/>        

	<descrption document="jjkk" />

	<part PartTitle="测试" PartNum="5" PartOver="1">

	    <sco ScoTitle="S001" isok="测试"  />

	    <quest Title="测试" />

	</part>

	<part PartTitle="测试" PartNum="5" PartOver="1">

	   	<sco ScoTitle="S001" isok="测试"  />

	    <quest Title="测试" />

		<!--间隔线-->

	    <sco ScoTitle="S001" isok="测试"  />

	    <quest Title="测试" />

	</part>

	<part PartTitle="测试"  PartNum="5" PartOver="1">

	    <sco ScoTitle="S001" isok="测试"  />

	    <quest Title="测试" />

	    <!--间隔线-->

	    <sco ScoTitle="S001" isok="测试"  />

	    <quest Title="测试" />

	    <!--间隔线-->

		<sco ScoTitle="S001" isok="测试"  />

		<quest Title="测试" />

	</part>

	....part节点

</teacher>

想得到teacher对象，大致类结构（设计的可能不合理）如下：

public class teacher

	{

		public string LessonTitle { get; set; }

		public general general{ get; set;}

		public descrption descrption{ get; set;}

		public List<part> partlist{ get; set; }

	}



	public class general

	{

		public string title { get; set; }

		public string msg { get; set; }

	}

	public class descrption

	{

		public string document { get; set; }

	}



	//================>part

	public class part

	{

		public string PartTitle { get; set; }

		public string PartNum { get; set; }

		public string PartOver { get; set; }

		List<kk> sco_quest{ get; set; }

	}

	public class kk

	{

		public sco sco{ get; set; }

		public quest quest{ get; set; }

	}

	public class sco

	{

		public string ScoTitle { get; set; }

		public string isok { get; set; }			

	}

	public class quest

	{

		public string Title { get; set; }

	}

如果是用XmlDocument来做可以做到，但是代码量有点多，如何快速用linq处理得到teacher对象呢。求支招