Problem: Snuke has found two trees A and B, each with N vertices numbered 1 to N. The i-th edge of A connects Vertex a_i and b_i, and the j-th edge of B connects Vertex c_j and d_j. Also, all vertices of A are initially painted white.

Snuke would like to perform the following operation on A zero or more times so that A coincides with B:

* Choose a leaf vertex that is painted white. (Let this vertex be v.)
* Remove the edge incident to v, and add a new edge that connects v to another vertex.
* Paint v black.



Determine if A can be made to coincide with B, ignoring color. If the answer is yes, find the minimum number of operations required.

You are given T cases of this kind. Find the answer for each of them.

Constraints

* 1 \leq T \leq 20
* 3 \leq N \leq 50
* 1 \leq a_i,b_i,c_i,d_i \leq N
* All given graphs are trees.

Input

Input is given from Standard Input in the following format:


T
case_1
:
case_{T}


Each case is given in the following format:


N
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 d_1
:
c_{N-1} d_{N-1}


Output

For each case, if A can be made to coincide with B ignoring color, print the minimum number of operations required, and print `-1` if it cannot.

Examples

Input

2
3
1 2
2 3
1 3
3 2
6
1 2
2 3
3 4
4 5
5 6
1 2
2 4
4 3
3 5
5 6


Output

1
-1


Input

3
8
2 7
4 8
8 6
7 1
7 3
5 7
7 8
4 2
5 2
1 2
8 1
3 2
2 6
2 7
4
1 2
2 3
3 4
3 4
2 1
3 2
9
5 3
4 3
9 3
6 8
2 3
1 3
3 8
1 7
4 1
2 8
9 6
3 6
3 5
1 8
9 7
1 6


Output

6
0
7

Solution (in CPP):

Solution:
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
const int INF=1e9;
int n;
void Clear(vector<int>& vec)
{
	vector<int>tmp;
	swap(vec,tmp);
	return ;
}
struct Tree{
	vector<int>to[110];
	void build(int u,int v)
	{
		to[u].push_back(v);
		return ;
	}
	void adde(int u,int v)
	{
		build(u,v);
		build(v,u);
		return ;
	}
	int fa[110];
	void clear()
	{
		for(int i=1;i<=n;i++)
		{
			Clear(to[i]);
			fa[i]=0;
		}
		return ;	
	}
	void dfs(int u,int lst=0)
	{
		fa[u]=lst;
		for(int i=0;i<to[u].size();i++)
			if(to[u][i]!=lst)dfs(to[u][i],u);
		return ;
	}
}TA,TB;
struct Graph{
	vector<int>to[110];
	int deg[110];
	void build(int u,int v)
	{
		to[u].push_back(v);
		deg[v]++;
		return ;
	}
	void clear()
	{
		for(int i=1;i<=n;i++)
		{
			Clear(to[i]);
			deg[i]=0;
		}
		return ;
	}
	bool check()
	{
		queue<int>q;
		for(int i=1;i<=n;i++)
			if(!deg[i])q.push(i);
		while(!q.empty())
		{
			int u=q.front();
			q.pop();
			for(int i=0;i<to[u].size();i++)
			{
				int v=to[u][i];
				deg[v]--;
				if(!deg[v])q.push(v);
			}
		}
		for(int i=1;i<=n;i++)
			if(deg[i])return false;
		return true;
	}
}G;
bool need(int u)
{
	return TA.fa[u]!=TB.fa[u];
}
int U[110],V[110];
int Deg[110];
int work(int u,int v)
{
	TA.clear();
	TA.adde(u,v);
	for(int i=1;i<n;i++)
	{
		if(U[i]==u||V[i]==u)continue;
		TA.adde(U[i],V[i]);
	}
	TA.dfs(u);TB.dfs(u);
	int res=1;
	for(int i=1;i<=n;i++)
	{
		res+=need(i);
		if(!TA.fa[i])continue;
		if(need(TA.fa[i])&&!need(i))
			return INF;
	}
	G.clear();
	for(int i=1;i<=n;i++)
	{
		if(TA.fa[i]){
			if(need(i)&&need(TA.fa[i]))
				G.build(i,TA.fa[i]);
		}
		if(TB.fa[i]){
			if(need(i)&&need(TB.fa[i]))
				G.build(TB.fa[i],i);
		}
	}
	if(G.check())return res;
	return INF;
}
void solve()
{
	scanf("%d",&n);
	TA.clear();TB.clear();
	for(int i=1;i<=n;i++)Deg[i]=0;
	for(int i=1;i<n;i++)
	{
		scanf("%d %d",&U[i],&V[i]);
		Deg[U[i]]++;Deg[V[i]]++;
		TA.adde(U[i],V[i]);
	}
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		TB.adde(u,v);
	}
	TA.dfs(1);TB.dfs(1);
	bool flag=false;
	for(int i=1;i<=n;i++)
		if(need(i))flag=true;
	if(!flag){
		printf("0\n");
		return ;
	}
	int ans=INF;
	for(int i=1;i<=n;i++)
	{
		if(Deg[i]!=1)continue;
		for(int j=1;j<=n;j++)
			if(i!=j)ans=min(ans,work(i,j));
	}
	if(ans>n)printf("-1\n");
	else printf("%d\n",ans);
	return ;
}
int main()
{
	int t;scanf("%d",&t);
	while(t--)
		solve();
	return 0;
}