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<bits/stdc++.h>
#define INF 1e9
using namespace std;
struct Info{int nu,ne;}a1[110],a2[110],a3[110];
int t,n,ansn,va[53],vb[53],ua[53],ub[53],f[53],b1[53],b2[53],b3[53],num1,num2,num3,su,p[53];
bool vi[53][53],fi[53];
void jb1(int x,int y){a1[++num1].nu=y;a1[num1].ne=b1[x];b1[x]=num1;}
void jb2(int x,int y){a2[++num2].nu=y;a2[num2].ne=b2[x];b2[x]=num2;}
void jb3(int x,int y){a3[++num3].nu=y;a3[num3].ne=b3[x];b3[x]=num3;}
void dfs(int x,int fa){
	fi[x]=true;su++;
	for (int y=b2[x];y;y=a2[y].ne)
		if (a2[y].nu!=fa&&vi[x][a2[y].nu])dfs(a2[y].nu,x);
}
void dfs1(int x,int fa){
	for (int y=b1[x];y;y=a1[y].ne){
		if (a1[y].nu!=fa){
			if (!fi[x])jb3(x,a1[y].nu);
			dfs1(a1[y].nu,x);
		}
	}
}
void dfs2(int x,int fa){
	for (int y=b2[x];y;y=a2[y].ne){
		if (a2[y].nu!=fa){
			if (!fi[x])jb3(a2[y].nu,x);
			dfs2(a2[y].nu,x);
		}
	}
}
bool dfs3(int x,int co){
	p[x]=co;
	for (int y=b3[x];y;y=a3[y].ne){
		if (p[a3[y].nu]==0){
			if (!dfs3(a3[y].nu,co))return false;
		}else{
			if (p[a3[y].nu]==co)return false;
		}
	}
	p[x]=n+1;
	return true;
}
bool pan(){
	int l=0;
	for (int i=1;i<=n;i++)p[i]=0;
	for (int i=1;i<=n;i++){
		if (p[i]==0){
			if (!dfs3(i,++l))return false;
		}
	}
	return true;
}
int work(int x){
	for (int i=1;i<=n;i++)b1[i]=0,b2[i]=0,b3[i]=0;num1=num2=num3=su=0;
	for (int i=1;i<n;i++){jb1(va[i],ua[i]);jb1(ua[i],va[i]);}
	for (int i=1;i<n;i++){vi[va[i]][ua[i]]=true;vi[ua[i]][va[i]]=true;}
	for (int i=1;i<n;i++){jb2(vb[i],ub[i]);jb2(ub[i],vb[i]);}
	for (int i=1;i<=n;i++) fi[i]=false;
	dfs(x,0);dfs1(x,0);dfs2(x,0);
	for (int i=1;i<n;i++){vi[va[i]][ua[i]]=false;vi[ua[i]][va[i]]=false;}
	if (pan()){return n-su;}return INF;
}
int main(){
	cin>>t;
	while (t--){
		cin>>n;ansn=INF;memset(f,0,sizeof(f));memset(vi,false,sizeof(vi));
		for (int i=1;i<n;i++){scanf("%d %d",&va[i],&ua[i]);f[va[i]]++;f[ua[i]]++;}
		for (int i=1;i<n;i++)scanf("%d %d",&vb[i],&ub[i]);
		for (int i=1;i<=n;i++)ansn=min(ansn,work(i));
		for (int i=1;i<=n;i++)
			if (f[i]==1)for (int j=1;j<=n;j++)
				if (i!=j)for (int k=1;k<=n;k++){
						if (va[k]==i){int p=ua[k];ua[k]=j;ansn=min(ansn,work(i)+1);ua[k]=p;}
						if (ua[k]==i){int p=va[k];va[k]=j;ansn=min(ansn,work(i)+1);va[k]=p;}
					}
		if (ansn==INF) cout<<"-1"<<endl;else cout<<ansn<<endl;
	}
	return 0;
}
/*
1
7
1 2
1 3
2 4
2 5
3 6
3 7
1 2
2 3
3 4
4 5
2 6
3 7
*/