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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct graph{
	struct edge{
		int s,t,nxt;
	}e[105];
	int last[55],e_cnt;
	void ins(int a,int b){
		e[e_cnt]=(edge){a,b,last[a]};
		last[a]=e_cnt++;
	}
	void clr(){
		memset(last,-1,sizeof(last));
		e_cnt=0;
	}
	bool vis[55],inst[55];
	int fa[55];
	void dfs(int x){
		for(int i=last[x];i!=-1;i=e[i].nxt)if(e[i].t!=fa[x]&&!vis[e[i].t]){
			fa[e[i].t]=x;
			dfs(e[i].t);
		}
	}
	bool chk(int x){
		if(inst[x])return false;
		if(vis[x])return true;
		vis[x]=1;
		inst[x]=1;
		for(int i=last[x];i!=-1;i=e[i].nxt)if(!chk(e[i].t))return false;
		inst[x]=0;
		return true;
	}
}A,B,C;
#define v(x) (A.fa[x]==B.fa[x])
int main(){
	// freopen("a.in","r",stdin);
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		A.clr();
		for(int i=1;i<n;i++){
			int u,v;
			scanf("%d%d",&u,&v);
			A.ins(u,v);
			A.ins(v,u);
		}
		B.clr();
		for(int i=1;i<n;i++){
			int u,v;
			scanf("%d%d",&u,&v);
			B.ins(u,v);
			B.ins(v,u);
		}
		int ans=1e9;
		for(int i=1;i<=n;i++){
			memset(A.fa,0,sizeof(A.fa));
			memset(B.fa,0,sizeof(B.fa));
			memset(A.vis,0,sizeof(A.vis));
			A.fa[i]=B.fa[i]=0;
			A.dfs(i); B.dfs(i);
			bool ok=1;
			int res=n-1;
			C.clr();
			for(int j=1;j<=n;j++)if(j!=i){
				if(v(j)){
					res--;
					if(!v(A.fa[j]))ok=0;
				}
				else{
					if(!v(A.fa[j]))C.ins(A.fa[j],j);
					if(!v(B.fa[j]))C.ins(j,B.fa[j]);
				}
			}
			if(!ok)continue;
			for(int j=1;j<=n;j++){
				memset(C.vis,0,sizeof(C.vis));
				memset(C.inst,0,sizeof(C.inst));
				if(!C.chk(j)){ok=0; break;}
			}
			if(ok)ans=min(ans,res);
		}
		if(ans<n)printf("%d\n",ans);
		else{
			for(int i=1;i<=n;i++)if(A.e[A.last[i]].nxt==-1)
				for(int t=1;t<=n;t++)if(t!=i){
					memset(A.vis,0,sizeof(A.vis));
					memset(A.fa,0,sizeof(A.fa));
					memset(B.fa,0,sizeof(B.fa));
					A.vis[i]=1;
					A.fa[i]=B.fa[i]=0;
					A.fa[t]=i;
					A.dfs(t); B.dfs(i);
					bool ok=1;
					int res=n-1;
					C.clr();
					for(int j=1;j<=n;j++)if(j!=i){
						if(v(j)){
							res--;
							if(!v(A.fa[j]))ok=0;
						}
						else{
							if(!v(A.fa[j]))C.ins(A.fa[j],j);
							if(!v(B.fa[j]))C.ins(j,B.fa[j]);
						}
					}
					if(!ok)continue;
					for(int j=1;j<=n;j++){
						memset(C.vis,0,sizeof(C.vis));
						memset(C.inst,0,sizeof(C.inst));
						if(!C.chk(j)){ok=0; break;}
					}
					if(ok)ans=min(ans,res+1);				
				}
			if(ans<=n)printf("%d\n",ans);
			else puts("-1");
		}
	}
	return 0;
}