Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

InputThere are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

OutputFor each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

Sample Input

3 3 31 2 31 2 31 2 331410

Sample Output

Case 1:NOYESNO 给三组整数，在下面每给一个整数，判断从三组数各取一个相加与此整数是否可以相等。 直接暴力复杂度为O(N^3)，不可行。可以将前两组枚举，复杂度为O(N^2)，再排序，对于每一个目标数，枚举第三组数，二分查找前两组的和即可。 源代码： #include
    
      #include
     
       #include
      
        using namespace std; int a[505],b[505],c[505],d[250005]; int main() { int L,M,N,S,s=1,x,y,z,flag; while(cin>>L>>M>>N) { y=0; flag=0; cout<<"Case "<
       
        <<":"<
        
         >a[i]; for(int i=0;i
         
          >b[i]; for(int i=0;i
          
           >c[i]; for(int j=0;j
           
            >S; for(int i=0;i
            
             >x; flag=0; for(int j=0;j