Code chef:divide The solution to the Tangerine Orange chunking algorithm

Once Chef decided to divide the tangerine into several parts. At the numbered Tangerine ' s segments from1ToNIn the clockwise order starting from some segment. Then He intended to divide the fruit into several parts. In order to does it he planned to separate the neighbouring segments inkPlaces, so and he could getkParts:the1^St -From segmentl₁ To segmentr₁ (inclusive), the2^nd -FromL₂ Tor₂ ,..., thek^th -Froml_k Tor_K (In the "cases in the" clockwise order). Suddenly, when Chef is absent, one naughty boy came and divided the tangerine intoPParts (also by separating the neighbouring segments one from another): the1^St -From segmenta₁ To segmentb₁ , the2^nd -Froma₂ Tob₂ ,..., thep^th -Froma_p Tob_P (In the "cases in the" clockwise order). Chef became very angry about it! But maybe little boy haven ' t done anything wrong, maybe everything are OK? Please, help Chef to determine whether him able to obtain the parts him wanted to have (in order to does it he can dividePCurrent parts, but, of the course, he can ' t join several parts into one.

Please, this is parts are not cyclic. That's means that even if the Tangerine Division consists of only one part, but "part" include more than one segment, the Re are two segments which were neighbouring in the initial tangerine, but not are in the neighbouring. The explanation of example case 2 to ensure your understood that clarification.

Input

The ' The ' input contains an integer T denoting the number of test cases. The description of T test cases follows.
The "a" of each test case contains three spaces 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 l and R_I. The next p lines contain pairs of space-separated integers a_i and b_I.

It is guaranteed so each tangerine ' s segment be contained in exactly one of the the 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-is AB Le to obtain the parts him wanted to have.

Constraints

1 ≤ T ≤

1 ≤ n ≤ 5 *⁷

1 ≤ k ≤ min (+, N)

1 ≤ p ≤ min (+, N)

1 ≤ l_i, r_I, a_i, b_i ≤ n

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Example

Input:
2
3 2 1 4 5 5 6 (1 5 6) 3 1 2 5 1 6 9 1

output:
   
Yes
No
  

Explanation

Example Case 1: To achieve his goal Chef should divide the "I" (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 is (though you may thought this way), he separated segments 1 and a from Ano Ther. But segments 1 and are in one part of Chef ' s division, so it's unable to achieve his goal.

Good long topic, the difficulty of this topic is how to read the topic.

Topic Resolution:

1 a piece of orange, and then the heap, all the heap of the fast number must be clockwise count

The 2-point heap is messed up.

3 in the chaos of the heap in the original want to divide the heap?

Additional hidden conditions: 1 All heaps need to be divided 2 can not be combined heap, can only separate heap

You can answer the questions by reading them, especially the hidden conditions.

Attention:

1 output format Yes not Yes

2 vector containers need to be zeroed every time.

The time efficiency of this program is somewhat complicated: 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[i].left)) return "No";  
return "Yes";  
    } 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