code chef

hongleimi

　　

Once Chef decided to divide the tangerine into several parts. At first, he numbered tangerine's segments from1 to n in
the clockwise order starting from some segment. Then he intended to divide the fruit into several parts. In order to do it he planned to separate the neighbouring segments in k places, so that he could get kparts: the 1st -
from segment l1 to segment r1 (inclusive), the 2nd - from l2 to r2, ..., the kth - from lk to rk (in
all cases in the clockwise order). Suddenly, when Chef was absent, one naughty boy came and divided the tangerine into p parts (also by separating the neighbouring segments one from another): the 1st - from segment a1 to
segment b1, the 2nd - from a2 to b2, ..., the pth - from ap to bp (in
all cases in the clockwise order). Chef became very angry about it! But maybe little boy haven't done anything wrong, maybe everything is OK? Please, help Chef to determine whether he is able to obtain the parts he wanted to have (in order to do it he can
divide p current parts, but, of course, he can't join several parts into one).

Please, note that parts are not cyclic. That means that even if the tangerine division consists of only one part, but that part include more than one segment, there are two segments which were neighbouring in the initial tangerine but are not neighbouring in
the division. See the explanation of example case 2 to ensure you understood that clarification.


Input


The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.

The first line of each test case contains three space separated integers n, k, p, denoting the number of tangerine's segments and number of parts in each of the two divisions. The next k lines
contain pairs of space-separated integers li and ri. The next p lines contain pairs of space-separated integers ai and bi.

It is guaranteed that each tangerine's segment is contained in exactly one of the first k parts and in exactly one of the next p parts.


Output


For each test case, output a single line containing either "Yes" or "No" (without the quotes), denoting whether Chef is able to obtain the parts he wanted to have.




Constraints

• 1 ≤ T ≤ 100
  
• 1 ≤ n ≤ 5 * 107
  
• 1 ≤ k ≤ min(500, n)
  
• 1 ≤ p ≤ min(500, n)
  
• 1 ≤ li, ri, ai, bi ≤ n
  

Example

Input:
2
10 3 2
1 4
5 5
6 10
1 5
6 10
10 3 1
2 5
10 1
6 9
1 10
Output:
Yes
No





Explanation


Example case 1: To achieve his goal Chef should divide the first part (1-5) in two by separating segments 4 and 5 one from another.

Example case 2: The boy didn't left the tangerine as it was (though you may thought that way), he separated segments 1 and 10 one from another. But segments 1 and 10 are in one part in Chef's division, so he is unable to achieve his goal.


　　好长的题目，本题目的难度就是如何读懂题目了。
　　题目解析：
　　1 把一个橘子分块，然后分堆，所有堆中有的快的序号必须是顺时针数起
　　2 分堆已经乱了
　　3 是否在乱了的堆中分出原来想要分的堆？
　　额外隐藏条件： 1 所有堆都需要分出来 2 不能合并堆，只能分开堆
　　读懂题目就能出答案了，尤其是隐藏条件
　　注意：
　　1 输出格式Yes不是YES  
　　2 vector容器每次都需要清零
　　这个程序的时间效率有点复杂，是：O(lg(k)*max(k, p))
　　#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
struct Tangerine
{
int left, right;
bool operator<(const Tangerine &t) const
{
return left < t.left;
}
};
bool biDiv(vector<Tangerine> &vk, int low, int up, int left)
{
if (low > up) return false;
int mid = low + ((up-low)>>1);
if (vk[mid].left < left) return biDiv(vk, mid+1, up, left);
if (left < vk[mid].left) return biDiv(vk, low, mid-1, left);
return true;
}
string divTangerine(vector<Tangerine> &vk, vector<Tangerine> &vp)
{
sort(vk.begin(), vk.end());
for (unsigned i = 0; i < vp.size(); i++)
{
if (!biDiv(vk, 0, vk.size()-1, vp.left)) return &quot;No&quot;;
}
return &quot;Yes&quot;;
}
void DivideTheTangerine()
{
int n = 0, k = 0, p = 0;
Tangerine tang;
vector<Tangerine> vk;
vector<Tangerine> vp;
int T = 0;
cin>>T;
while (T--)
{
cin>>n>>k>>p;
for (int i = 0; i < k; i++)
{
cin>>tang.left>>tang.right;
vk.push_back(tang);
}
for (int i = 0; i < p; i++)
{
cin>>tang.left>>tang.right;
vp.push_back(tang);
}
cout<<divTangerine(vk, vp)<<endl;
vk.clear(), vp.clear();
}
}

OJ: http://www.codechef.com/problems/TANGDIV