#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#define PB push_back
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=55;
int n;
namespace Gr {
	vector<int> G[N];
	queue<int> que;
	int du[N];
	void init() { for(int i=1;i<=n;++i) G[i].clear(),du[i]=0; }
	void ad(int x,int y) {
//		cout<<x<<' '<<y<<endl;
		G[x].PB(y);
		du[y]++;
	}
	bool sol() {
		int cnt=0;
		for(int i=1;i<=n;++i) if(!du[i]) que.push(i);
		while(!que.empty()) {
			cnt++; int u=que.front(); que.pop();
			for(int i=0;i<G[u].size();++i) {
				int v=G[u][i];
				if(!(--du[v])) que.push(v);
			}
		}
		return cnt==n;
	}
}
vector<int> G[2][N];
int fa[2][N];
int mp[2][N][N];
int flg;
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;
}
void dfs(int p,int u,int last) {
	fa[p][u]=last;
	for(int i=0;i<G[p][u].size();++i)
		if(G[p][u][i]!=last) dfs(p,G[p][u][i],u);
}
int vis[N],tot_siz;
void check(int u) {
	if(vis[u]) return; tot_siz++,vis[u]=1;
	for(int i=0;i<G[0][u].size();++i) check(G[0][u][i]);
}
int sol(int rt) {
//	cout<<"SOL:"<<rt<<endl;
	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].PB(v);
		}
	{
		tot_siz=0;
		for(int i=1;i<=n;++i) vis[i]=0;
		check(1);
		if(tot_siz<n) return 0;
	}		
		
	dfs(0,rt,0);
	dfs(1,rt,0);
	Gr::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]]) Gr::ad(fa[1][i],i);
			if(fa[1][fa[0][i]]!=fa[0][fa[0][i]]) Gr::ad(i,fa[0][i]);
		}
		else cnt++;
	if(dfs2(rt)!=cnt) return 0;
	
	if(Gr::sol()) {
//		cout<<"OK"<<endl;
		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; rd(T);
	while(T--) {
		flg=0;
		rd(n);
		for(int i=1,x,y;i<n;++i) rd(x),rd(y),mp[0][x][y]=mp[0][y][x]=1;
		for(int i=1,x,y;i<n;++i) rd(x),rd(y),mp[1][x][y]=mp[1][y][x]=1;
		int ans=sol(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,sol(u)-1);
					mp[0][u][v]=mp[0][v][u]=1;
					mp[0][u][w]=mp[0][w][u]=0;
				}
		}
		if(!flg) printf("-1\n");
		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;
	}
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#define PB push_back
#define ll long long
using namespace std;
template <class T>
inline void rd(T &x) {
	x=0; char c=getchar(); int f=1;
	while(!isdigit(c)) { if(c=='-') f=-1; c=getchar(); }
	while(isdigit(c)) x=x*10-'0'+c,c=getchar(); x*=f;
}
const int N=55;
int n;
namespace Gr {
	vector<int> G[N];
	queue<int> que;
	int du[N];
	void init() { for(int i=1;i<=n;++i) G[i].clear(),du[i]=0; }
	void ad(int x,int y) {
//		cout<<x<<' '<<y<<endl;
		G[x].PB(y);
		du[y]++;
	}
	bool sol() {
		int cnt=0;
		for(int i=1;i<=n;++i) if(!du[i]) que.push(i);
		while(!que.empty()) {
			cnt++; int u=que.front(); que.pop();
			for(int i=0;i<G[u].size();++i) {
				int v=G[u][i];
				if(!(--du[v])) que.push(v);
			}
		}
		return cnt==n;
	}
}
vector<int> G[2][N];
int fa[2][N];
int mp[2][N][N];
int flg;
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;
}
void dfs(int p,int u,int last) {
	fa[p][u]=last;
	for(int i=0;i<G[p][u].size();++i)
		if(G[p][u][i]!=last) dfs(p,G[p][u][i],u);
}
int vis[N],tot_siz;
void check(int u) {
	if(vis[u]) return; tot_siz++,vis[u]=1;
	for(int i=0;i<G[0][u].size();++i) check(G[0][u][i]);
}
int sol(int rt) {
//	cout<<"SOL:"<<rt<<endl;
	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].PB(v);
		}
	{
		tot_siz=0;
		for(int i=1;i<=n;++i) vis[i]=0;
		check(1);
		if(tot_siz<n) return 0;
	}		
		
	dfs(0,rt,0);
	dfs(1,rt,0);
	Gr::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]]) Gr::ad(fa[1][i],i);
			if(fa[1][fa[0][i]]!=fa[0][fa[0][i]]) Gr::ad(i,fa[0][i]);
		}
		else cnt++;
	if(dfs2(rt)!=cnt) return 0;
	
	if(Gr::sol()) {
//		cout<<"OK"<<endl;
		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; rd(T);
	while(T--) {
		flg=0;
		rd(n);
		for(int i=1,x,y;i<n;++i) rd(x),rd(y),mp[0][x][y]=mp[0][y][x]=1;
		for(int i=1,x,y;i<n;++i) rd(x),rd(y),mp[1][x][y]=mp[1][y][x]=1;
		int ans=sol(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,sol(u)-1);
					mp[0][u][v]=mp[0][v][u]=1;
					mp[0][u][w]=mp[0][w][u]=0;
				}
		}
		if(!flg) printf("-1\n");
		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;
	}
	return 0;
}
```