// Problem: P3919 【模板】可持久化线段树 1（可持久化数组）
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P3919
// Memory Limit: 1 MB
// Time Limit: 1500 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int, int> pii;
#define x first
#define y second
const int N = 1e5 + 10;
int a[N], root[N];
int n, m, k;
struct Persistable_Segment_Tree {
    int lc[N * 30], rc[N * 30], val[N * 30], cnt;
    inline void build(int& o, int l, int r) {
        o = ++cnt;
        if (l == r) { return; }
        int mid = (l + r) >> 1;
        build(lc[o], l, mid); build(rc[o], mid + 1, r);
    }
    inline void ins(int& o, int pre, int l, int r, int q) {
        o = ++cnt; lc[o] = lc[pre]; rc[o] = rc[pre]; val[o] = val[pre] + 1;
        if (l == r) { return; }
        int mid = (l + r) >> 1;
        if (q <= mid)ins(lc[o], lc[pre], l, mid, q);
        else ins(rc[o], rc[pre], mid + 1, r, q);
    }
    inline int query(int o, int l, int r, int ql, int qr) {
    	if(ql > qr) return 0;
    	if(ql <= l && qr >= r){
    		return val[o];
    	}
        int mid = (l + r) >> 1, res = 0;
        if (ql <= mid) res += query(lc[o], l, mid, ql, qr);
        if (qr > mid) res += query(rc[o], mid + 1, r, ql, qr);
        return res;
    }
    
    inline void clear(){
    	for(int i = 0; i <= cnt; i++){
    		lc[i] = rc[i] = val[i] = 0;
    	}
    	cnt = 0;
    }
}T;

map<int, int> mp;
multiset<int> maxv;

void split(int l, int r, int x){
	int base = mp[l - 1]; // [l ~ r]整个区间的逆序对
	maxv.erase(maxv.lower_bound(base));
	int t1 = 0, t2 = 0; 
	// t1是 (l ~ x) 或者 (x ~ r) 内的逆序对, 
	// t2是跨了两个区间的逆序对
	// cout << l << ' ' << r << ' ' << x << '\n'; 
 	if(x - l < r - x){
		for(int i = l + 1; i <= x; i++){
			t1 += T.query(root[i - 1], 1, n, a[i] + 1, n)
				- T.query(root[l - 1], 1, n, a[i] + 1, n);
		}
		for(int i = l; i <= x; i++){
			t2 += T.query(root[r], 1, n, 1, a[i] - 1)
				- T.query(root[x], 1, n, 1, a[i] - 1);
		}
		
		int temp = T.query(root[x - 1], 1, n, a[x] + 1, n)
			- T.query(root[l - 1], 1, n, a[x] + 1, n);
		
		mp[l - 1] = t1 - temp, mp[x] = base - t1 - t2;
		maxv.insert(mp[l - 1]); maxv.insert(mp[x]);
	}
	else{
		  for(int i = x + 1; i <= r; i++){
		  	t1 += T.query(root[i - 1], 1, n, a[i] + 1, n)
		  		- T.query(root[x - 1], 1, n, a[i] + 1, n);
		  }
		  for(int i = x; i <= r; i++){
		  	t2 += T.query(root[x - 1], 1, n, a[i] + 1, n)
		  		- T.query(root[l - 1], 1, n, a[i] + 1, n);
		  }
		  
		  int temp = T.query(root[r], 1, n, 1, a[x] - 1)
		  	- T.query(root[x], 1, n, 1, a[x] - 1);
		  mp[x] = t1 - temp, mp[l - 1] = base - t1 - t2;
		  maxv.insert(mp[x]); maxv.insert(mp[l - 1]);
	}
}
void solve(){
	mp.clear(); maxv.clear();
	cin >> n;
	for(int i = 1; i <= n; i++){
		cin >> a[i];
	}
	T.build(root[0], 1, n);
	for(int i = 1; i <= n; i++){
		T.ins(root[i], root[i - 1], 1, n, a[i]);
	}
	
	int base = 0;
	for(int i = 2; i <= n; i++){
		base += T.query(root[i - 1], 1, n, a[i] + 1, n);
	}
	mp[0] = base, mp[n + 1] = 0;
	maxv.insert(base);
	
	int ans = *maxv.rbegin();
	for(int i = 1; i <= n; i++){
		cout << ans << " \n"[i == n];
		int x; cin >> x;
		x ^= ans;
		auto it = mp.lower_bound(x);
		split((prev(it)->x) + 1, (it->x) - 1, x);
		ans = *maxv.rbegin();
		
	}
	T.clear();
}
signed main(){
    std::ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int t = 1;
    cin >> t;
    while(t--){
        solve();
    }
    return 0;
}