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

Solution (in CPP):

Solution:
#include<iostream>
#include<iomanip>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int read()
{
	char c=getchar();
	int res=0;
	while(c<'0'||c>'9') c=getchar();
	while(c>='0'&&c<='9') res=(res<<1)+(res<<3)+(c^48),c=getchar();
	return res;
}
struct Edge
{
	int to,next;
}w[1005];
int t,n,d[55],d2[55],pre[55],sum,rd[55],h[55],cnt,prt[55];
bool flags=1;
bool ma[55][55],ma2[55][55];
bool vis[55],vst[55];
void add(int x,int y)
{
	w[++cnt]=(Edge){y,h[x]};
	h[x]=cnt;rd[y]++;
}
void dfs(int x,int fa)
{
	vis[x]=1;prt[x]=fa;
	for(int i=1;i<=n;i++) if(ma2[x][i]&&!vis[i]) dfs(i,x);
}
void dfs2(int x,int fa)
{
	vis[x]=1;
	if(prt[x]!=fa) vst[x]=1,sum++;
	else vst[x]=0;
	if(fa) add(fa,x);
	int s=1;
	for(int i=1;i<=n;i++) if(ma[x][i]&&!vis[i]) dfs2(i,x),s&=vst[i];
	if(vst[x]&&!s) flags=0;
}
void dfs3(int x,int fa)
{
	vis[x]=1;
	if(vst[x]&&vst[fa]) add(x,fa);
	for(int i=1;i<=n;i++) if(ma2[x][i]&&!vis[i]) dfs3(i,x);
}
bool solve(int x)
{
	memset(rd,0,sizeof(rd));
	memset(vis,0,sizeof(vis));memset(h,0,sizeof(h));cnt=0;
	dfs(x,0);
	memset(vis,0,sizeof(vis));
	flags=1;
	dfs2(x,0);
	memset(vis,0,sizeof(vis));
	dfs3(x,0);
	if(!flags) return 0;
	queue<int> q;
	for(int i=1;i<=n;i++) if(!rd[i]) q.push(i);
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		for(int i=h[u];i;i=w[i].next)
		{
			int y=w[i].to;
			if(!(--rd[y])) q.push(y);
		}
	}
	
	for(int i=1;i<=n;i++) if(rd[i]) return 0;
	return 1;
}
int main()
{
	t=read();
	while(t--)
	{
		memset(ma,0,sizeof(ma));memset(ma2,0,sizeof(ma2));
		memset(d,0,sizeof(d));
		n=read();
		for(int i=1;i<n;i++)
		{
			int x,y;
			x=read(),y=read();
			d[x]++,d[y]++;
			pre[x]=y;pre[y]=x;
			ma[x][y]=ma[y][x]=1;
		}
		for(int i=1;i<n;i++)
		{
			int x,y;
			x=read(),y=read();
			ma2[x][y]=ma2[y][x]=1;
		}
		bool flag=1;
		for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++) if(ma[i][j]^ma2[i][j]) flag=0;
		if(flag)
		{
			puts("0");
			continue;
		}
		int ans=1e9;
		for(int u=1;u<=n;u++)
		{
			sum=0;
			if(solve(u)) ans=min(ans,sum);
			if(d[u]>1) continue;
			for(int v=1;v<=n;v++)
			{
				if(u==v) continue;
				if(v==pre[u]) continue;
				sum=1;
				ma[u][pre[u]]=ma[pre[u]][u]=0;
				ma[u][v]=ma[v][u]=1;
				if(solve(u)) ans=min(ans,sum);
				ma[u][v]=ma[v][u]=0;
				ma[u][pre[u]]=ma[pre[u]][u]=1;
			}
		}
		if(ans==1e9) puts("-1");
		else printf("%d\n",ans);
	}
}