#include <bits/stdc++.h>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
#define ll long long
//#define file
using namespace std;

int a[51][51],b[51][51],c[51],d[51],T,n,i,j,k,l,ans,t;
bool bz1[51],bz2[51],Bz[51],Bz2[51];

void add(int t) {if (bz1[t] && bz2[t] && !Bz[t]) d[++::t]=t,Bz[t]=1;}
void bfs(int st)
{
	static int d[51];
	int i,h=0,t=1;
	d[1]=st;
	while (h<t)
	{
		++h;Bz2[d[h]]=1;
		fo(i,1,n)
		if (a[d[h]][i] && b[d[h]][i] && !Bz[i])
		Bz[i]=1,d[++t]=i,::d[++::t]=i;
	}
}
void work(int st,int Ans)
{
	static int c[51];
	int i,j,k,l;
	
	memset(bz1,0,sizeof(bz1));memset(bz2,0,sizeof(bz2));
	memset(Bz,0,sizeof(Bz));memset(Bz2,0,sizeof(Bz2));
	memset(c,0,sizeof(c));
	t=1;d[1]=st;Bz[st]=1;bfs(st);
	fo(i,1,n-1)
	{
		fo(j,i+1,n)
		if (a[i][j])
		++c[i],++c[j];
	}
	fo(i,1,n) if (c[i]==1) bz1[i]=1;
	Ans+=n-t;
	
	while (t)
	{
		l=d[t],--t;
		if (!Bz2[l])
		{
			Bz2[l]=1;
			fo(i,1,n)
			if (a[l][i])
			{
				--c[i];
				if (c[i]==1) bz1[i]=1,add(i);
			}
		}
		fo(i,1,n) if (b[l][i]) bz2[i]=1,add(i);
	}
	
	fo(i,1,n) if (!Bz[i]) return;
	ans=min(ans,Ans);
}

int main()
{
	#ifdef file
	freopen("agc027f.in","r",stdin);
	freopen("b.out","w",stdout);
	#endif
	
	scanf("%d",&T);
	for (;T;--T)
	{
		memset(a,0,sizeof(a)),memset(b,0,sizeof(b)),memset(c,0,sizeof(c));
		scanf("%d",&n);
		fo(i,1,n-1) scanf("%d%d",&j,&k),a[j][k]=a[k][j]=1,++c[j],++c[k];
		fo(i,1,n-1) scanf("%d%d",&j,&k),b[j][k]=b[k][j]=1;
		
		ans=114514;
		fo(i,1,n)
		if (c[i]==1)
		{
			fo(j,1,n) if (a[i][j]) break;
			fo(k,1,n)
			if (!a[i][k] && i!=k && j!=k)
			{
				a[i][j]=a[j][i]=0,a[i][k]=a[k][i]=1;
				work(i,1);
				a[i][j]=a[j][i]=1,a[i][k]=a[k][i]=0;
			}
			work(i,0);
		}
		printf("%d\n",(ans>n)?-1:ans);
	}
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}
### Prompt

Your task is to create a Cpp solution to 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>
#define fo(a,b,c) for (a=b; a<=c; a++)
#define fd(a,b,c) for (a=b; a>=c; a--)
#define min(a,b) (a<b?a:b)
#define ll long long
//#define file
using namespace std;

int a[51][51],b[51][51],c[51],d[51],T,n,i,j,k,l,ans,t;
bool bz1[51],bz2[51],Bz[51],Bz2[51];

void add(int t) {if (bz1[t] && bz2[t] && !Bz[t]) d[++::t]=t,Bz[t]=1;}
void bfs(int st)
{
	static int d[51];
	int i,h=0,t=1;
	d[1]=st;
	while (h<t)
	{
		++h;Bz2[d[h]]=1;
		fo(i,1,n)
		if (a[d[h]][i] && b[d[h]][i] && !Bz[i])
		Bz[i]=1,d[++t]=i,::d[++::t]=i;
	}
}
void work(int st,int Ans)
{
	static int c[51];
	int i,j,k,l;
	
	memset(bz1,0,sizeof(bz1));memset(bz2,0,sizeof(bz2));
	memset(Bz,0,sizeof(Bz));memset(Bz2,0,sizeof(Bz2));
	memset(c,0,sizeof(c));
	t=1;d[1]=st;Bz[st]=1;bfs(st);
	fo(i,1,n-1)
	{
		fo(j,i+1,n)
		if (a[i][j])
		++c[i],++c[j];
	}
	fo(i,1,n) if (c[i]==1) bz1[i]=1;
	Ans+=n-t;
	
	while (t)
	{
		l=d[t],--t;
		if (!Bz2[l])
		{
			Bz2[l]=1;
			fo(i,1,n)
			if (a[l][i])
			{
				--c[i];
				if (c[i]==1) bz1[i]=1,add(i);
			}
		}
		fo(i,1,n) if (b[l][i]) bz2[i]=1,add(i);
	}
	
	fo(i,1,n) if (!Bz[i]) return;
	ans=min(ans,Ans);
}

int main()
{
	#ifdef file
	freopen("agc027f.in","r",stdin);
	freopen("b.out","w",stdout);
	#endif
	
	scanf("%d",&T);
	for (;T;--T)
	{
		memset(a,0,sizeof(a)),memset(b,0,sizeof(b)),memset(c,0,sizeof(c));
		scanf("%d",&n);
		fo(i,1,n-1) scanf("%d%d",&j,&k),a[j][k]=a[k][j]=1,++c[j],++c[k];
		fo(i,1,n-1) scanf("%d%d",&j,&k),b[j][k]=b[k][j]=1;
		
		ans=114514;
		fo(i,1,n)
		if (c[i]==1)
		{
			fo(j,1,n) if (a[i][j]) break;
			fo(k,1,n)
			if (!a[i][k] && i!=k && j!=k)
			{
				a[i][j]=a[j][i]=0,a[i][k]=a[k][i]=1;
				work(i,1);
				a[i][j]=a[j][i]=1,a[i][k]=a[k][i]=0;
			}
			work(i,0);
		}
		printf("%d\n",(ans>n)?-1:ans);
	}
	
	fclose(stdin);
	fclose(stdout);
	return 0;
}
```