#include<bits/stdc++.h>
using namespace std;

int n,T,x,y,ans,d[61],fa[61],col[61],Col[61],dd[61],nw;
vector<int> v[61],V[61];

void addedge(int x,int y) {v[x].push_back(y),v[y].push_back(x);}
void Addedge(int x,int y) {V[x].push_back(y),V[y].push_back(x);}

bool work()
{
	for (int i=1; i<nw; i++)
	{
		int p=0;
		for (int j=1; j<=n; j++) if (col[j]==1&&Col[j]==1) {p=j; break;}
		if (!p)
		{
			memset(col,0,sizeof(col));
			memset(Col,0,sizeof(Col));
			return 0;
		}
		col[p]=Col[p]=2;
		for (int j=0,sz=v[p].size(); j<sz; j++)
			if ((--dd[v[p][j]])==1) if (!col[v[p][j]]) col[v[p][j]]=1;
		for (int j=0,sz=V[p].size(); j<sz; j++)
			if (V[p][j]!=fa[p]) Col[V[p][j]]=1;
	}
	memset(col,0,sizeof(col));
	memset(Col,0,sizeof(Col));
	return 1;
}

void getfa(int x)
{
	for (int i=0,sz=V[x].size(); i<sz; i++)
		if (V[x][i]!=fa[x]) fa[V[x][i]]=x,getfa(V[x][i]);
}

void dfs(int x)
{
	col[x]=Col[x]=2,nw--;
	for (int i=0,sz=V[x].size(); i<sz; i++)
		if (V[x][i]!=fa[x])
		{
			bool bo=0;
			for (int k=0,siz=v[x].size(); k<siz; k++)
				if (v[x][k]==V[x][i]) {bo=1; break;}
			if (bo) dfs(V[x][i]); else Col[V[x][i]]=1;
		}
}

int main()
{
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n),ans=n+1;
		for (int i=1; i<=n; i++) v[i].clear(),V[i].clear(),d[i]=0;
		for (int i=1; i<n; i++) scanf("%d%d",&x,&y),addedge(x,y),d[x]++,d[y]++;
		for (int i=1; i<n; i++) scanf("%d%d",&x,&y),Addedge(x,y);
		if (n<=2) {puts("0"); continue;}
		for (int i=1; i<=n; i++)
		{
			Col[i]=1,fa[i]=0,getfa(i),nw=n;
			for (int j=1; j<=n; j++) if ((dd[j]=d[j])==1) col[j]=1;
			if (work()) {ans=n; break;}
		}
		for (int i=1; i<=n; i++)
		{
			nw=n,fa[i]=0,getfa(i),dfs(i);
			for (int j=1; j<=n; j++) if ((dd[j]=d[j])==1) if (!col[j]) col[j]=1;
			if (work()) ans=min(ans,nw);
		}
		if (ans==n+1) puts("-1"); else printf("%d\n",ans);
	}
	return 0;
}
### Prompt

Create a solution in CPP for the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;

int n,T,x,y,ans,d[61],fa[61],col[61],Col[61],dd[61],nw;
vector<int> v[61],V[61];

void addedge(int x,int y) {v[x].push_back(y),v[y].push_back(x);}
void Addedge(int x,int y) {V[x].push_back(y),V[y].push_back(x);}

bool work()
{
	for (int i=1; i<nw; i++)
	{
		int p=0;
		for (int j=1; j<=n; j++) if (col[j]==1&&Col[j]==1) {p=j; break;}
		if (!p)
		{
			memset(col,0,sizeof(col));
			memset(Col,0,sizeof(Col));
			return 0;
		}
		col[p]=Col[p]=2;
		for (int j=0,sz=v[p].size(); j<sz; j++)
			if ((--dd[v[p][j]])==1) if (!col[v[p][j]]) col[v[p][j]]=1;
		for (int j=0,sz=V[p].size(); j<sz; j++)
			if (V[p][j]!=fa[p]) Col[V[p][j]]=1;
	}
	memset(col,0,sizeof(col));
	memset(Col,0,sizeof(Col));
	return 1;
}

void getfa(int x)
{
	for (int i=0,sz=V[x].size(); i<sz; i++)
		if (V[x][i]!=fa[x]) fa[V[x][i]]=x,getfa(V[x][i]);
}

void dfs(int x)
{
	col[x]=Col[x]=2,nw--;
	for (int i=0,sz=V[x].size(); i<sz; i++)
		if (V[x][i]!=fa[x])
		{
			bool bo=0;
			for (int k=0,siz=v[x].size(); k<siz; k++)
				if (v[x][k]==V[x][i]) {bo=1; break;}
			if (bo) dfs(V[x][i]); else Col[V[x][i]]=1;
		}
}

int main()
{
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d",&n),ans=n+1;
		for (int i=1; i<=n; i++) v[i].clear(),V[i].clear(),d[i]=0;
		for (int i=1; i<n; i++) scanf("%d%d",&x,&y),addedge(x,y),d[x]++,d[y]++;
		for (int i=1; i<n; i++) scanf("%d%d",&x,&y),Addedge(x,y);
		if (n<=2) {puts("0"); continue;}
		for (int i=1; i<=n; i++)
		{
			Col[i]=1,fa[i]=0,getfa(i),nw=n;
			for (int j=1; j<=n; j++) if ((dd[j]=d[j])==1) col[j]=1;
			if (work()) {ans=n; break;}
		}
		for (int i=1; i<=n; i++)
		{
			nw=n,fa[i]=0,getfa(i),dfs(i);
			for (int j=1; j<=n; j++) if ((dd[j]=d[j])==1) if (!col[j]) col[j]=1;
			if (work()) ans=min(ans,nw);
		}
		if (ans==n+1) puts("-1"); else printf("%d\n",ans);
	}
	return 0;
}
```