Can you find it?

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. 

Input

There 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. 

Output

For 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 3
1 2 3
1 2 3
1 2 3
3
1
4
10

Sample Output

Case 1:
NO
YES
NO

#include <iostream>
#include   <cstring>
#include   <cstdlib>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include <algorithm>
using namespace std;
long long  x,v;
int l,n,m,s;
long long  L[555],N[555],M[555],LN[500*500+500];
bool sreach(long long  k){
    long long  li=1;
    long long  r=v;
    while(li<=r){
        long long  mid=(li+r)/2;
        if(LN[mid]==k)return 1;
        else if(LN[mid]<k)
            li=mid+1;
        else
            r=mid-1;
    }

    return 0;
}
int main()
{
    int t=0;
    while(scanf("%d%d%d",&l,&n,&m)!=EOF){
        for(int i=1;i<=l;i++)
            scanf("%lld",&L[i]);
        for(int i=1;i<=n;i++)
            scanf("%lld",&N[i]);
        for(int i=1;i<=m;i++)
            scanf("%lld",&M[i]);
        v=0;
        for(int i=1;i<=l;i++){
            for(int j=1;j<=n;j++){
               LN[++v]=L[i]+N[j];
               //cout << LN[v] << endl;
            }
        }
        scanf("%d",&s);
        printf("Case %d:\n",++t);

        sort(LN+1,LN+v+1);
        while(s--){
            scanf("%lld",&x);
            int flag=0;
            for(int i=1;i<=m;i++){

                if(sreach(x-M[i])){
                    flag=1;
                    break;
                }

            }
            if(flag)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    //cout << "Hello world!" << endl;
    return 0;
}