/* 
 * Problem: POI2001 pch
 * Author: Guo Jiabao
 * Time: 2009.2.5 14:05
 * State: Solved 
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
using namespace std;
const int MAXN=4001,PERN=2000;
struct edge{int t;edge *next;};
struct sequence{int len;string s;};
struct subsq{int len;string *s;};
template <typename T> class linknode
{
	public:
		linknode *next;
		T s;
		linknode(T ts):next(0),s(ts){}
};
template <typename T> class linklist
{
	public:
		linknode<T> *f,*l;
		linklist():f(0),l(0){}
		void ins(T s)
		{
			if (l)l=l->next=new linknode<T>(s);
			else f=l=new linknode<T>(s);
		}
		void clear()
		{
			linknode<T> *a=f,*b;
			while (a){b=a;a=a->next;delete b;}
			f=l=0;
		}
};
int D,N,Ec,Bc,S1,S2;
int Bel[MAXN];//每個頂點所屬連通分量
int TB[MAXN],TC; //暫存正向搜索的連通分量
linklist<int> Circle[MAXN]; //鏈表存儲每個圈
bool inCircle[MAXN]; //記錄每個節節點
bool used[MAXN]; //生成排列時DFS用到的記錄數組
sequence seq[MAXN]; //每個子樹的括號序列
int Cs[MAXN]; //每個圈的大小
edge E[MAXN*2];//邊集的空間
edge *Ap[MAXN],*An[MAXN];//正圖和逆圖的鄰接表
subsq S[MAXN];//暫存括號序列

void addedge(edge *&A,int t)
{
	if (A)
	{
		edge *t=A;
		A=&E[++Ec];
		A->next=t;
	}
	else
	{
		A=&E[++Ec];
		A->next=0;
	}
	E[Ec].t=t;
}

void init()
{
	int i,t;
	Ec=-1;Bc=0;
	memset(Ap,0,sizeof(Ap));
	memset(An,0,sizeof(An));
	scanf("%d",&N);
	for (i=1;i<=N;i++)
	{
		scanf("%d",&t);
		addedge(Ap[i],t);
		addedge(An[t],i);
		Bel[i]=0;
		seq[i].len=0;
		seq[i].s="";
		inCircle[i]=false;
	}
	for (i=PERN+1;i<=PERN+N;i++)
	{
		scanf("%d",&t);
		t+=PERN;
		addedge(Ap[i],t);
		addedge(An[t],i);
		Bel[i]=0;
		seq[i].len=0;
		seq[i].s="";
		inCircle[i]=false;
	}
}

void connect_dfs_n(int u) //逆向求連通分量
{
	Bel[u]=Bc;
	for (edge *k=An[u];k;k=k->next)
	{
		int v=k->t;
		if (!Bel[v])
			connect_dfs_n(v);
	}
}

int connect_dfs_p(int u) //正向求連通分量，並查找圈
{
	Bel[u]=Bc;
	TB[++TC]=u;
	int v=Ap[u]->t;
	if (!Bel[v])
	{
		int r=connect_dfs_p(v);
		if (r)
		{
			++Cs[Bc];
			Circle[Bc].ins(u);
			inCircle[u]=true;
			if (r==u)
				return 0;
			else
				return r;
		}
		else
			return 0;
	}
	else
	{
		++Cs[Bc];
		Circle[Bc].ins(u);
		inCircle[u]=true;
		if (u==v) v=0;
		return v;
	}
}

void connect(int A,int B)
{
	for (int i=A;i<=B;i++)
	{
		if (!Bel[i])
		{
			Bc++;
			Cs[Bc]=0;
			TC=0;
			connect_dfs_p(i);
			for (int j=1;j<=TC;j++)
				connect_dfs_n(TB[j]);
		}
	}
}

inline int cmp(const void *a,const void *b)
{
	return  ((subsq *)a)->s->compare(*(((subsq *)b)->s)) ;
}

void TreeSeq(int u)//求內向樹序列
{
	linklist<subsq> LL;
	subsq T;
	int L=0;
	++seq[u].len;
	seq[u].s+='(';
	for (edge *k=An[u];k;k=k->next)
	{
		int v=k->t;
		if (inCircle[v]) continue;
		TreeSeq(v);
		T.s=&seq[v].s;
		T.len=seq[v].len;
		LL.ins(T);
	}
	for (linknode<subsq> *p=LL.f;p;p=p->next)
		S[++L]=p->s;
	qsort(S+1,L,sizeof(S[0]),cmp);
	for (int i=1;i<=L;i++)
	{
		seq[u].s+=*(S[i].s);
		seq[u].len+=S[i].len;
	}
	++seq[u].len;
	seq[u].s+=')';
	LL.clear();
}

bool compare(int a,int b)
{
	if (Cs[a]!=Cs[b]) return false; //圈大小不相等
	linknode<int> *k,*l,*pl;
	for (pl=Circle[b].f;pl;pl=pl->next) //枚舉偏移量
	{
		l=pl;
		for (k=Circle[a].f;k;k=k->next)
		{
			int x=k->s,y=l->s;
			if (!(l=l->next)) l=Circle[b].f;
			if (seq[x].len!=seq[y].len || seq[x].s!=seq[y].s) //內向樹序列不相同
				break;
		}
		if (!k)
			return true;
	}
	return false;
}

bool arrange_dfs(int k) //生成兩圖連通分量對應關係的排列
{
	if (k>S1)
		return true;
	for (int i=S1+1;i<=S2;i++)
		if (!used[i] && compare(k,i))
		{
			used[i]=true;
			bool b=arrange_dfs(k+1);
			used[i]=false;
			if (b) return true;
		}
	return false;
}

bool solve()
{
	int i,j;
	init();
	connect(1,N);S1=Bc;
	connect(1+PERN,N+PERN);S2=Bc;
	if (S1!=S2-S1) //1..S1爲圖1的連通分量 S1+1..S2爲圖2的連通分量
		return false; //兩圖連通分量個數不相等
	for (i=1;i<=S2;i++)
		for (linknode<int> *k=Circle[i].f;k;k=k->next)
			TreeSeq(k->s); //計算以每個圈上頂點爲根的內向樹括號序列
	return arrange_dfs(1);
}

void clear()
{
	for (int i=1;i<=S2;i++)
		Circle[i].clear();
}

int main()
{
	int i;
	freopen("pch.in","r",stdin);
	freopen("pch.out","w",stdout);
	scanf("%d",&D);
	for (i=1;i<=D;i++)
	{
		if (solve())
			printf("Tn");
		else
			printf("Nn");
		clear(); //清理 釋放內存
	}
	return 0;
}