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

//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]; pIl tmp[mx]; //b prefsums
I f2(int l, int r, int t, ll B, I c)
{
    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; //first number that begins with 1
    if((c>>t)&1) return f2(m,r,t-1,B,c-(I(1)<<t));
    bool b1=b[r]-b[m]<=B;
    if(b1) return f2(l,m,t-1,B-(b[r]-b[m]),c)+(I(1)<<t);
    return max(f2(l,m,t-1,B,c),f2(m,r,t-1,B,c-(I(1)<<t)));
}
I f1(int l, int r, int t, ll B)
{
    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; //first number that begins with 1
    bool b0=b[m]-b[l]<=B, b1=b[r]-b[m]<=B;
    if(!b0&&b1) return f1(l,m,t-1,B-(b[r]-b[m]))+(I(1)<<t);
    if(!b0&&!b1) return max(f1(l,m,t-1,B),f1(m,r,t-1,B));
    if(b1&&l==m) return (a[l]&((I(1)<<(t+1))-1))+(I(1)<<(t+1))-1;
    I ans = f2(m,r,t-1,B-(b[m]-b[l]),(I(1)<<t)-(a[m-1]&((I(1)<<(t+1))-1)))+(I(1)<<t);
    if(!b1&&l<m) return ans;
    return max(ans,f1(l,m,t-1,B-(b[r]-b[m])));
}
I solve()
{
    int n, k; ll m; cin>>n>>m>>k;
    //split nums in 2 groups based on highest bit
    //if can add for all 1s but not all 0s: x gets 1
        //must add for 1s group -> discard, recurse on 0s
    //if can add for all 0s but not all 1s: x gets 0 WLOG
        //must add for 0s group (*)
    //if cannot add for either group: answer highest bit is 0
        //discard numbers with different top bit from x: recurse twice
    //if can add for either group:
        //if x gets 1: add for 1s group, recurse on 0s if any
        //if x gets 0: add for 0s group, by (*)
    //(*) solve same problem with additional term x-c (c>0) so add = delete
    //if c has top bit 1: x>=c gets 1, recurse on 1s
    //if c has 0 and can add for all 1s: x gets 1 and answer can get 1
    //if c has 0 and cannot add: answer has 0, recurse twice
    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()
{
    cin.sync_with_stdio(0), cin.tie(0), cout.sync_with_stdio(0), cout.tie(0);
    int T; cin>>T; while(T--) write(solve()), cout<<'\n';
}
//?