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<bits/stdc++.h>
using namespace std;
const int N = 55;
int n;

struct Tree{
	vector<int> g[N];
	int deg[N];
	queue<int> q;
	inline void init(){
		for(int i=1;i<=n;i++)g[i].clear(), deg[i]=0;
	}
	inline void adde(int u,int v){
		g[u].push_back(v),deg[v]++;
	}
	inline bool solve(){
		int cnt=0;
		for(int i=1;i<=n;i++)if(!deg[i])q.push(i);
		while(!q.empty()){
			++cnt;int u=q.front();q.pop();
			for(size_t i=0;i<g[u].size();i++){
				int v=g[u][i];
				--deg[v];if(!deg[v])q.push(v);
			}
		}
		return cnt==n;
	}
}dag;

vector<int> g[2][N];
int fa[2][N];
int mp[2][N][N];
int flg;

inline int dfs2(int u){
	int tot=1;for(int v=1;v<=n;v++)if(fa[0][v]==u&&fa[1][v]==u){
		tot+=dfs2(v);
	}
	return tot;
}

inline void dfs(int p,int u,int pre){
	fa[p][u]=pre;
	for(size_t i=0;i<g[p][u].size();i++){
		int v=g[p][u][i];if(v==pre)continue;
		dfs(p,v,u);
	}
}

int vis[N],tsiz;
inline void check(int u){
	if(vis[u])return; tsiz++, vis[u]=1;
	for(size_t i=0;i<g[0][u].size();i++)check(g[0][u][i]);
}

inline int solve(int rt){
	
	
	for(int d=0;d<2;++d){
		for(int u=1;u<=n;u++){
			g[d][u].clear();
			for(int v=1;v<=n;v++)if(mp[d][u][v])g[d][u].push_back(v);
		}
	}
		tsiz=0;
		for(int i=1;i<=n;i++)vis[i]=0;
		check(1);
		if(tsiz<n)return 0;
	dfs(0,rt,0);
	dfs(1,rt,0);
	dag.init();
	int cnt=0;
	for(int i=1;i<=n;i++){
		if(fa[0][i]!=fa[1][i]){
			if(fa[1][fa[1][i]]!=fa[0][fa[1][i]]) dag.adde(fa[1][i],i);
			if(fa[1][fa[0][i]]!=fa[0][fa[0][i]]) dag.adde(i, fa[0][i]);
		}
		else cnt++;
	}
	if(dfs2(rt)!=cnt)return 0;
	if(dag.solve()){
		flg=1;int cnt=0;
		for(int i=1;i<=n;++i)cnt+=fa[1][i]==fa[0][i];
		return cnt;
	}
	return 0;
}

int main()
{
	int T;cin >> T;
	while(T--){
		flg=0;
		scanf("%d",&n);
		for(int i=1,x,y;i<n;i++){scanf("%d%d",&x,&y);mp[0][x][y]=mp[0][y][x]=1;}
		for(int i=1,x,y;i<n;i++){scanf("%d%d",&x,&y);mp[1][x][y]=mp[1][y][x]=1;}
		int ans = solve(1);
		for(int u=1;u<=n;u++){
			int t=0;for(int v=1;v<=n;v++)t+=mp[0][u][v];
			if(t>1)continue;
			for(int v=1;v<=n;v++)if(mp[0][u][v]){
				for(int w=1;w<=n;w++)if(!mp[0][u][w]){
					
					mp[0][u][v]=mp[0][v][u]=0;
					mp[0][u][w]=mp[0][w][u]=1;
					
					ans = max(ans, solve(u)-1);
					mp[0][u][v]=mp[0][v][u]=1;
					mp[0][u][w]=mp[0][w][u]=0;
				}
			}
		}
		if(!flg){
			puts("-1");
		}
		else {
			printf("%d\n",n-ans);
		}
		for(int d=0;d<2;d++)
			for(int i=1;i<=n;i++)
				for(int j=1;j<=n;j++)
					mp[d][i][j]=0;
	}
}