#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int N=55;
int T,n,du[N],foo,bar,fa1[N],fa2[N],mrk[N],vis[N],ans;
vector<int>E1[N],E2[N],G[N];
queue<int>Q;
void dfs2(int u,int f){
	fa2[u]=f;
	for(int v:E2[u])
		if(v!=f)
			dfs2(v,u);
}
void dfs1(int u,int f){
	fa1[u]=f;mrk[u]=mrk[f]&(fa1[u]==fa2[u]);
	for(int v:E1[u])
		if(v!=f&&(u!=foo||v!=bar)&&(v!=foo||u!=bar))
			dfs1(v,u);
}
int calc(int rt){
	dfs2(rt,0);mrk[0]=1;dfs1(rt,0);
	for(int i=1;i<=n;++i)G[i].clear(),vis[i]=0;
	int res=0,tmp=0;
	for(int i=1;i<=n;++i)
		if(!mrk[i]){
			++res;
			if(!mrk[fa1[i]])
				G[i].push_back(fa1[i]),++vis[fa1[i]];
			if(!mrk[fa2[i]])
				G[fa2[i]].push_back(i),++vis[i];
		}
	while(!Q.empty())Q.pop();
	for(int i=1;i<=n;++i)
		if(!mrk[i]&&!vis[i])
			Q.push(i);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		++tmp;
		for(int v:G[u])
			if(!--vis[v])
				Q.push(v);
	}
	return tmp==res?res:1<<30;
}
int main(){
	scanf("%d",&T);while(T--){
		scanf("%d",&n);
		for(int i=1;i<=n;++i)E1[i].clear(),E2[i].clear(),du[i]=0;
		for(int i=1,x,y;i<n;++i){
			scanf("%d%d",&x,&y);
			E1[x].push_back(y);
			E1[y].push_back(x);
			++du[x];++du[y];
		}
		for(int i=1,x,y;i<n;++i){
			scanf("%d%d",&x,&y);
			E2[x].push_back(y);
			E2[y].push_back(x);
		}
		foo=bar=0;ans=calc(1);
		for(int i=1;i<=n;++i)
			if(du[i]==1){
				int j=E1[i][0];
				foo=i;bar=j;
				for(int k=1;k<=n;++k)
					if(k!=i&&k!=j){
						E1[i].push_back(k);
						E1[k].push_back(i);
						ans=min(ans,calc(i)+1);
						E1[i].pop_back();
						E1[k].pop_back();
					}
			}
		printf("%d\n",ans==1<<30?-1:ans);
	}
	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<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int N=55;
int T,n,du[N],foo,bar,fa1[N],fa2[N],mrk[N],vis[N],ans;
vector<int>E1[N],E2[N],G[N];
queue<int>Q;
void dfs2(int u,int f){
	fa2[u]=f;
	for(int v:E2[u])
		if(v!=f)
			dfs2(v,u);
}
void dfs1(int u,int f){
	fa1[u]=f;mrk[u]=mrk[f]&(fa1[u]==fa2[u]);
	for(int v:E1[u])
		if(v!=f&&(u!=foo||v!=bar)&&(v!=foo||u!=bar))
			dfs1(v,u);
}
int calc(int rt){
	dfs2(rt,0);mrk[0]=1;dfs1(rt,0);
	for(int i=1;i<=n;++i)G[i].clear(),vis[i]=0;
	int res=0,tmp=0;
	for(int i=1;i<=n;++i)
		if(!mrk[i]){
			++res;
			if(!mrk[fa1[i]])
				G[i].push_back(fa1[i]),++vis[fa1[i]];
			if(!mrk[fa2[i]])
				G[fa2[i]].push_back(i),++vis[i];
		}
	while(!Q.empty())Q.pop();
	for(int i=1;i<=n;++i)
		if(!mrk[i]&&!vis[i])
			Q.push(i);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		++tmp;
		for(int v:G[u])
			if(!--vis[v])
				Q.push(v);
	}
	return tmp==res?res:1<<30;
}
int main(){
	scanf("%d",&T);while(T--){
		scanf("%d",&n);
		for(int i=1;i<=n;++i)E1[i].clear(),E2[i].clear(),du[i]=0;
		for(int i=1,x,y;i<n;++i){
			scanf("%d%d",&x,&y);
			E1[x].push_back(y);
			E1[y].push_back(x);
			++du[x];++du[y];
		}
		for(int i=1,x,y;i<n;++i){
			scanf("%d%d",&x,&y);
			E2[x].push_back(y);
			E2[y].push_back(x);
		}
		foo=bar=0;ans=calc(1);
		for(int i=1;i<=n;++i)
			if(du[i]==1){
				int j=E1[i][0];
				foo=i;bar=j;
				for(int k=1;k<=n;++k)
					if(k!=i&&k!=j){
						E1[i].push_back(k);
						E1[k].push_back(i);
						ans=min(ans,calc(i)+1);
						E1[i].pop_back();
						E1[k].pop_back();
					}
			}
		printf("%d\n",ans==1<<30?-1:ans);
	}
	return 0;
}

```