#pragma GCC optimize ("O3")
#include "bits/stdc++.h"
using namespace std;

#define rep(i, b, e) for(int i = (b); i <= (e); i++)
#define per(i, b, e) for(int i = (e); i >= (b); i--)
#define FOR(i, b, e) rep(i, b, (e) - 1)
#define SZ(x) int(x.size())
#define all(x) x.begin(), x.end()
#define pb push_back
#define mp make_pair
#define st first
#define nd second
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

auto &operator<<(auto &o, pair<auto, auto> p) {
    return o << "(" << p.st << ", " << p.nd << ")";
}

auto operator<<(auto &o, auto x)->decltype(end(x), o) {
    o << "{"; int i=0; for(auto e: x) o << ", " + 2*!i++ << e;
    return o << "}";
}

#ifdef LOCAL
#define deb(x...) cerr << "[" #x << "]: ", [](auto...$) { ((cerr << $ << "; "),...) << endl; }(x)
#else
#define deb(...)
#endif

void solve() {
    int n;
    cin>>n;
    vi A(n), B(n);
    for(int &i:A)cin>>i;
    for(int &i:B)cin>>i;
    sort(all(A));
    sort(all(B));
    int ans=0;
    for(int i=0; i<=n; i++){//tyle najm z a parujemy z b
        int akt=2e9;
        for(int j=0; j<i; j++){
            akt=min(akt, abs(B[n-i+j]-A[j]));
        }
        for(int j=i;j<n; j++){
            akt=min(akt, abs(B[j-i]-A[j]));
        }
        ans=max(ans, akt);
    }
    cout<<ans<<"\n";
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    int tt = 1;
    cin >> tt;
    FOR(te, 0, tt) solve();
    return 0;
}