#pragma GCC optimize("O3")
#include <iostream>
#include <algorithm>
using namespace std;
const int mx=100000;
typedef long long ll; typedef __int128 I;

//128io
void read(__int128 &x){
	// read a __int128 variable
	char c; bool f = 0;
	while(((c = getchar()) < '0' || c > '9') && c != '-');
	if(c == '-'){f = 1; c = getchar();}
	x = c - '0';
	while((c = getchar()) >= '0' && c <= '9')x = x * 10 + c - '0';
	if(f) x = -x;
}
void write(__int128 x){
	// print a __int128 variable
	if(x < 0){putchar('-'); x = -x;}
	if(x > 9)write(x / 10);
	putchar(x % 10 + '0');
}

I a[mx]; ll b[mx+1]; pair<I,ll> tmp[mx]; //b prefsums
I f1(int,int,int,ll);
I f2(int,int,int,ll,I);
I f3(int,int,int,ll,I);
I f4(int,int,int,ll);
I f1(int l, int r, int t, ll B)
{
// for(int i=0; i<9-t; i++) cerr<<'|'<<' ';
// cerr<<"1\t\tl="<<l<<" r="<<r<<" t="<<t<<" B="<<B<<'\n';
    if(t==-1) return 0;
    //add all
    if(b[r]-b[l]<=B) return (a[l]&((I(1)<<(t+1))-1))+(I(1)<<(t+1))-1;
    int L=l-1, R=r;
    while(R!=L+1) if((a[(L+R)/2]>>t)&1) R=(L+R)/2; else L=(L+R)/2;
    int m=R;
    if(l==m) return f1(l,r,t-1,B)+(I(1)<<t);
    bool b1=b[r]-b[m]<=B;
    if(b[m]-b[l]>B)
    {
        //!b0&&b1: x gets 1, 1s add, discard
        if(b1) return f1(l,m,t-1,B-(b[r]-b[m]))+(I(1)<<t);
        //neither: if x gets 0, discard 1s
        //if x gets 1, discard 0s, 1s add=del
        return max(f1(l,m,t-1,B),f4(m,r,t-1,B));
    }
    //have b0 so can achieve 2^t-1
    I ans = (I(1)<<t)-1;
    //if x gets 0, 0s add; assuming no actual 0
    I n1=a[l]&((I(1)<<(t+1))-1);
    if(n1) ans = f2(m,r,t-1,B-(b[m]-b[l]),(I(1)<<t)-n1)+(I(1)<<t);
    //if b1, can do !b0&&b1
    if(b1) ans = max(ans,f1(l,m,t-1,B-(b[r]-b[m]))+(I(1)<<t));
    return ans;
}
I f2(int l, int r, int t, ll B, I c)
{
// for(int i=0; i<9-t; i++) cerr<<'|'<<' ';
// cerr<<"2\t\tl="<<l<<" r="<<r<<" t="<<t<<" B="<<B<<" c="<<ll(c)<<'\n';
    //same problem with additional term x-c (0<=c<2^(t+1))
    //implicit add := delete
    if(t==-1) return 0; if(l==r) return (I(1)<<(t+1))-1-c;
    int L=l-1, R=r;
    while(R!=L+1) if((a[(L+R)/2]>>t)&1) R=(L+R)/2; else L=(L+R)/2;
    int m=R;
    //c has 1: x gets 1, delete 0s
    if((c>>t)&1) return f2(m,r,t-1,B,c-(I(1)<<t));
    //if can add for all 1s: x gets 1
    if(b[r]-b[m]<=B) return f2(l,m,t-1,B-(b[r]-b[m]),c)+(I(1)<<t);
    //if x gets 0: discard 1s
    //if x gets 1: discard 0s, now x+c and add := delete
    return max(f2(l,m,t-1,B,c),f3(m,r,t-1,B,(I(1)<<t)-c));
}
I f4(int l, int r, int t, ll B)
{
// for(int i=0; i<9-t; i++) cerr<<'|'<<' ';
// cerr<<"4\t\tl="<<l<<" r="<<r<<" t="<<t<<" B="<<B<<'\n';
    //same problem but add := delete; guaranteed cannot delete all
    if(t==-1) return 0;
    int L=l-1, R=r;
    while(R!=L+1) if((a[(L+R)/2]>>t)&1) R=(L+R)/2; else L=(L+R)/2;
    int m=R;
    //delete one group
    if(b[m]-b[l]<=B) return f4(m,r,t-1,B-(b[m]-b[l]))+(I(1)<<t);
    if(b[r]-b[m]<=B) return f4(l,m,t-1,B-(b[r]-b[m]))+(I(1)<<t);
    //neither
    return max(f4(l,m,t-1,B),f4(m,r,t-1,B));
}
I f3(int l, int r, int t, ll B, I c)
{
// for(int i=0; i<9-t; i++) cerr<<'|'<<' ';
// cerr<<"3\t\tl="<<l<<" r="<<r<<" t="<<t<<" B="<<B<<" c="<<ll(c)<<'\n';
    //same problem but add := delete; guaranteed cannot delete all
    //also one additional term x+c (0<=c<=2^(t+1))
    if(t==-1) return 0; if(b[r]-b[l]<=B) return (I(1)<<(t+1))-1+c;
    int L=l-1, R=r;
    while(R!=L+1) if((a[(L+R)/2]>>t)&1) R=(L+R)/2; else L=(L+R)/2;
    int m=R;
    //delete one group
    if(b[m]-b[l]<=B) return f3(m,r,t-1,B-(b[m]-b[l]),c)+(I(1)<<t);
    if(b[r]-b[m]<=B) return f3(l,m,t-1,B-(b[r]-b[m]),c)+(I(1)<<t);
    //neither
    return max(f3(l,m,t-1,B,c),f3(m,r,t-1,B,c));
}
I solve()
{
    int n, k; ll m; cin>>n>>m>>k;
    //compute answer starting with highest bit
    //also solve problem with extra term x-c
    for(int i=0; i<n; i++) read(tmp[i].first);
    for(int i=0; i<n; i++) cin>>tmp[i].second;
    sort(tmp,tmp+n);
    for(int i=0; i<n; i++) a[i]=tmp[i].first, b[i+1]=b[i]+tmp[i].second;
    return f1(0,n,k-1,m);
}
int main()
{
    int T; cin>>T>>T; while(T--) write(solve()), cout<<'\n';
}