#pragma GCC optimize("3","Ofast","inline")
#include<bits/stdc++.h>
#define _ 0
using namespace std;
typedef long long ll;
template<class T>inline void read(T &x) {
    x=0;T f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-48,ch=getchar();
    x*=f;
}
int T,n;
vector<int>G[55],F[55];
inline void dfs(int u,vector<int>Edg[55],vector<int>&f) {
	for(int i=0;i<(int)Edg[u].size();++i)
		if(Edg[u][i]!=f[u])
			f[Edg[u][i]]=u,dfs(Edg[u][i],Edg,f);
}
inline int calc(int root1,int root2) {
	vector<int>fa1(n+1,0),fa2(n+1,0),added(n+1,0),deg(n+1,0);
	dfs(root1,G,fa1),dfs(root2,F,fa2);
	added[0]=1;
	if(root1==root2) {
		for(int i=1;i<=n+1;++i)
			for(int j=1;j<=n;++j)
				if(added[fa1[j]]&&fa1[j]==fa2[j])
					added[j]=1;
	}
	for(int i=1;i<=n;++i)
		deg[fa1[i]]++;
	int res=0;
	for(int i=1;i<=n;++i) {
		int nxt=-1;
		for(int j=1;j<=n&&!~nxt;++j)
			if(!deg[j]&&!added[j]&&added[fa2[j]])
				nxt=j;
		if(!~nxt)
			break;
		added[nxt]=1;
		deg[fa1[nxt]]--;
		res++;
	}
	return !*min_element(added.begin(),added.end())?1e9:res;
}
int main() {
	read(T);
	while(T--) {
		for(int i=1;i<=50;++i)
			F[i].clear(),G[i].clear();
		read(n);
		for(int i=1,x,y;i<n;++i)
			read(x),read(y),G[x].push_back(y),G[y].push_back(x);
		for(int i=1,x,y;i<n;++i)
			read(x),read(y),F[x].push_back(y),F[y].push_back(x);
		int ans=1e9;
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				ans=min(ans,calc(i,j));
		printf("%d\n",ans==1e9?-1:ans);
	}
	return ~~(0^_^0);
}
### Prompt

Develop a solution in Cpp 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
#pragma GCC optimize("3","Ofast","inline")
#include<bits/stdc++.h>
#define _ 0
using namespace std;
typedef long long ll;
template<class T>inline void read(T &x) {
    x=0;T f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-48,ch=getchar();
    x*=f;
}
int T,n;
vector<int>G[55],F[55];
inline void dfs(int u,vector<int>Edg[55],vector<int>&f) {
	for(int i=0;i<(int)Edg[u].size();++i)
		if(Edg[u][i]!=f[u])
			f[Edg[u][i]]=u,dfs(Edg[u][i],Edg,f);
}
inline int calc(int root1,int root2) {
	vector<int>fa1(n+1,0),fa2(n+1,0),added(n+1,0),deg(n+1,0);
	dfs(root1,G,fa1),dfs(root2,F,fa2);
	added[0]=1;
	if(root1==root2) {
		for(int i=1;i<=n+1;++i)
			for(int j=1;j<=n;++j)
				if(added[fa1[j]]&&fa1[j]==fa2[j])
					added[j]=1;
	}
	for(int i=1;i<=n;++i)
		deg[fa1[i]]++;
	int res=0;
	for(int i=1;i<=n;++i) {
		int nxt=-1;
		for(int j=1;j<=n&&!~nxt;++j)
			if(!deg[j]&&!added[j]&&added[fa2[j]])
				nxt=j;
		if(!~nxt)
			break;
		added[nxt]=1;
		deg[fa1[nxt]]--;
		res++;
	}
	return !*min_element(added.begin(),added.end())?1e9:res;
}
int main() {
	read(T);
	while(T--) {
		for(int i=1;i<=50;++i)
			F[i].clear(),G[i].clear();
		read(n);
		for(int i=1,x,y;i<n;++i)
			read(x),read(y),G[x].push_back(y),G[y].push_back(x);
		for(int i=1,x,y;i<n;++i)
			read(x),read(y),F[x].push_back(y),F[y].push_back(x);
		int ans=1e9;
		for(int i=1;i<=n;++i)
			for(int j=1;j<=n;++j)
				ans=min(ans,calc(i,j));
		printf("%d\n",ans==1e9?-1:ans);
	}
	return ~~(0^_^0);
}
```