#include <bits/stdc++.h>
using namespace std;

using ll = long long;

const int MAXN = 1e5+5;

multiset<ll, greater<ll>> mset;

int T;
int n;
int a[MAXN];
ll p[MAXN];

struct BIT
{
    int t[MAXN];

    inline int lowbit(int x) {return x&(-x);}

    inline void clear() {memset(t, 0, sizeof(int)*(n+5));}

    void add(int x)
    {
        while(x <= n)
        {
            ++t[x];
            x += lowbit(x);
        }
    }

    int query(int x)
    {
        int ret = 0;
        while(x)
        {
            ret += t[x];
            x -= lowbit(x);
        }
        return ret;
    }
}bit;

ll getinvcnt()
{
    ll ret = 0;
    for(int i = n; i >= 1; --i)
    {
        ret += bit.query(a[i]-1);
        bit.add(a[i]);
    }
    return ret;
}

void myout(int id, ll ans)
{
    if(id == n) printf("%lld",ans);
    else printf("%lld ",ans);
}

struct node
{
    int l,r; ll invcnt;
    bool operator<(const node& ano) const
    {
        return l < ano.l;
    }
};

set<node> qujian;

struct hjttree
{
    struct hjtnode
    {
        int ls,rs,v;
    }t[MAXN<<5];

    #define mid ((l+r)>>1)

    int root[MAXN], cnt;

    void clear()
    {
        memset(t, 0, sizeof(hjtnode)*(cnt+2));
        cnt = 0;
    }

    void build(int &o, int pre, int l, int r, int x)
    {
        o = ++cnt;
        t[o] = t[pre];
        ++t[o].v;
        if(l == r) return;
        if(x <= mid) build(t[o].ls, t[pre].ls, l, mid, x);
        else build(t[o].rs, t[pre].rs, mid+1, r, x);
    }

    int querycnt(int l, int r, int L, int R)
    {
        if(L > R) return 0;
        return realcnt(root[r], root[l-1], 1, n, L, R);
    }

    //查询 ox - oy 有多少个在 [L,R] 的数
    int realcnt(int ox, int oy, int l, int r, int L, int R)
    {
        if(L <= l && r <= R) return t[ox].v - t[oy].v;
        int ret = 0;
        if(L <= mid) ret += realcnt(t[ox].ls, t[oy].ls, l, mid, L, R);
        if(R > mid) ret += realcnt(t[ox].rs, t[oy].rs, mid+1, r, L, R);
        return ret;
    }

}seg;

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        bit.clear(); qujian.clear(); seg.clear();
        for(int i = 1; i <= n; ++i)
        {
            scanf("%d",&a[i]);
            seg.build(seg.root[i], seg.root[i-1], 1, n, a[i]);
        }
        for(int i = 1; i <= n; ++i) scanf("%lld",&p[i]);
        ll invcnt = getinvcnt();
        qujian.insert({1, n, invcnt}); mset.insert(invcnt);
        myout(1, invcnt);
        ll ans = invcnt;
        for(int i = 1; i < n; ++i)
        {
            p[i] ^= ans;

            //按p[i]位置分裂
            auto it = qujian.upper_bound((node){p[i], 0, 0});
            --it;
            int l = it->l, r = it->r;

            ll invcntleft = 0, invcntright = 0, invcntcross = 0, invcntp = 0;
            //启发式处理
            if(p[i]-l <= r-p[i])        //左边短
            {
                for(int j = l; j <= p[i]-1; ++j)
                {
                    invcntleft += seg.querycnt(j+1, p[i]-1, 1, a[j]-1);
                    invcntcross += seg.querycnt(p[i]+1, r, 1, a[j]-1);
                }
                invcntp = seg.querycnt(l,p[i]-1,a[p[i]]+1,n) + seg.querycnt(p[i]+1,r,1,a[p[i]]-1);
                invcntright = it->invcnt - invcntleft - invcntcross - invcntp;
            }
            else                        //右边短
            {
                for(int j = p[i]+1; j <= r; ++j)
                {
                    invcntright += seg.querycnt(j+1, r, 1, a[j]-1);
                    invcntcross += seg.querycnt(l, p[i]-1, a[j]+1, n);
                }
                invcntp = seg.querycnt(l,p[i]-1,a[p[i]]+1,n) + seg.querycnt(p[i]+1,r,1,a[p[i]]-1);
                invcntleft = it->invcnt - invcntright - invcntcross - invcntp;
            }

            //最后要插回去区间
            mset.erase(mset.find(it->invcnt)); mset.insert(invcntleft); mset.insert(invcntright);
            qujian.erase(it);
            if(l <= p[i]-1) qujian.insert((node){l, p[i]-1, invcntleft});
            if(p[i]+1 <= r) qujian.insert((node){p[i]+1, r, invcntright});
            ans = *mset.begin();
            myout(i, ans);
        }
        if(T) printf("\n");
    }
    return 0;
}