#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
const int N=57;
int read(){int x;scanf("%d",&x);return x;}
std::vector<int>e[N],g[N];
int fa[N],ndeg[N],deg[N],vis[N],mp[N][N];
void dfs(int u,int f)
{
    if(fa[u]=f,vis[f]&&mp[u][f]) vis[u]=1;
    for(int v:g[u]) if(v^f) dfs(v,u);
}
void work()
{
    int n=read(),ans=n+1;memset(mp,0,sizeof mp);
    for(int i=1;i<=n;++i) e[i].clear(),g[i].clear();
    for(int i=1,u,v;i<n;++i) u=read(),v=read(),mp[u][v]=mp[v][u]=1,e[u].push_back(v),e[v].push_back(u);
    for(int i=1,u,v;i<n;++i) u=read(),v=read(),g[u].push_back(v),g[v].push_back(u);
    for(int i=1;i<=n;++i) ndeg[i]=e[i].size();
    for(int i=1;i<=n;++i)
    {
	int cost=0,u;
	memset(vis,0,4*n+4),vis[i]=1,dfs(i,0),memcpy(deg+1,ndeg+1,4*n);
	while(1)
	{
	    for(u=1;u<=n;++u) if(!vis[u]&&deg[u]==1&&vis[fa[u]]) break;
	    if(u>n) break; else vis[u]=1,++cost;
	    for(int v:e[u]) --deg[v];
	}
	for(u=1;u<=n;++u) if(!vis[u]) break;
	if(u>n) ans=std::min(ans,cost);
	if(e[i].size()>1||u>n||ans<=n) continue;
	memset(vis,0,4*n+4),vis[0]=1,memcpy(deg+1,ndeg+1,4*n);
	while(1)
	{
	    for(u=1;u<=n;++u) if(!vis[u]&&deg[u]<=1&&vis[fa[u]]) break;
	    if(u>n) break; else vis[u]=1;
	    for(int v:e[u]) --deg[v];
	}
	for(u=1;u<=n;++u) if(!vis[u]) break;
	if(u>n) ans=n;
    }
    printf("%d\n",ans>n? -1:ans);
}
int main(){for(int T=read();T;--T)work();}

### Prompt

Construct a cpp code solution to the problem outlined:

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<vector>
#include<cstring>
#include<algorithm>
const int N=57;
int read(){int x;scanf("%d",&x);return x;}
std::vector<int>e[N],g[N];
int fa[N],ndeg[N],deg[N],vis[N],mp[N][N];
void dfs(int u,int f)
{
    if(fa[u]=f,vis[f]&&mp[u][f]) vis[u]=1;
    for(int v:g[u]) if(v^f) dfs(v,u);
}
void work()
{
    int n=read(),ans=n+1;memset(mp,0,sizeof mp);
    for(int i=1;i<=n;++i) e[i].clear(),g[i].clear();
    for(int i=1,u,v;i<n;++i) u=read(),v=read(),mp[u][v]=mp[v][u]=1,e[u].push_back(v),e[v].push_back(u);
    for(int i=1,u,v;i<n;++i) u=read(),v=read(),g[u].push_back(v),g[v].push_back(u);
    for(int i=1;i<=n;++i) ndeg[i]=e[i].size();
    for(int i=1;i<=n;++i)
    {
	int cost=0,u;
	memset(vis,0,4*n+4),vis[i]=1,dfs(i,0),memcpy(deg+1,ndeg+1,4*n);
	while(1)
	{
	    for(u=1;u<=n;++u) if(!vis[u]&&deg[u]==1&&vis[fa[u]]) break;
	    if(u>n) break; else vis[u]=1,++cost;
	    for(int v:e[u]) --deg[v];
	}
	for(u=1;u<=n;++u) if(!vis[u]) break;
	if(u>n) ans=std::min(ans,cost);
	if(e[i].size()>1||u>n||ans<=n) continue;
	memset(vis,0,4*n+4),vis[0]=1,memcpy(deg+1,ndeg+1,4*n);
	while(1)
	{
	    for(u=1;u<=n;++u) if(!vis[u]&&deg[u]<=1&&vis[fa[u]]) break;
	    if(u>n) break; else vis[u]=1;
	    for(int v:e[u]) --deg[v];
	}
	for(u=1;u<=n;++u) if(!vis[u]) break;
	if(u>n) ans=n;
    }
    printf("%d\n",ans>n? -1:ans);
}
int main(){for(int T=read();T;--T)work();}

```