#include<bits/stdc++.h>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
int n;
struct Tree{
	set<int> e[110];
	int fa[110];
	void init(){
		for(int i=1;i<n;i++){
			int x,y;read(x);read(y);
			e[x].insert(y);
			e[y].insert(x);
		}
	}
	void dfs(int x,int dad){
		fa[x]=dad;
		for(int v:e[x]) if(v!=dad) dfs(v,x);
	}
	int solve(int *p,int rt){
		dfs(rt,0);
		int s=0,tag[110]={};
		for(int i=1;i<=n;i++)
			if(i!=rt) tag[i]=fa[i]!=p[i],s+=tag[i];
		for(int i=1;i<=n;i++)
			if(i!=rt&&tag[fa[i]]&&!tag[i]) return 1e9;
		int d[110]={};
		vector<int> pre[110];
		for(int i=1;i<=n;i++){
			if(i!=rt&&tag[i]&&tag[fa[i]]){
				pre[fa[i]].push_back(i);
				d[i]++;
			}
			if(i!=rt&&tag[i]&&tag[p[i]]){
				pre[i].push_back(p[i]);
				d[p[i]]++;
			}
		}
		queue<int> q;int now=0;
		for(int i=1;i<=n;i++)
			if(i!=rt&&tag[i]&&!d[i]) q.push(i);
		while(!q.empty()){
			int u=q.front();q.pop();now++;
			for(int v:pre[u])
				if(--d[v]==0) q.push(v);
		}
		if(now==s) return s;
		else return 1e9;
	}
};
int main(){
	int T;read(T);
	while(T--){
		Tree T1,T2;
		read(n);
		T1.init();
		T2.init();
		int flag=1,ans=1e9;
		for(int i=1;i<=n;i++)
			if(T1.e[i]!=T2.e[i]) flag=0;
		if(flag){puts("0");continue;}
		for(int i=1;i<=n;i++){
			T2.dfs(i,0);
			if(T1.e[i].size()!=1) continue;
			for(int j=1;j<=n;j++)
				if(j!=i){
					Tree t=T1;
					t.e[*t.e[i].begin()].erase(i);
					t.e[i].clear();
					t.e[i].insert(j);
					t.e[j].insert(i);
					ans=min(ans,t.solve(T2.fa,i)+1);
				}
		}
		if(ans>1e8) ans=-1;
		cout<<ans<<'\n';
	}
	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 ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
int n;
struct Tree{
	set<int> e[110];
	int fa[110];
	void init(){
		for(int i=1;i<n;i++){
			int x,y;read(x);read(y);
			e[x].insert(y);
			e[y].insert(x);
		}
	}
	void dfs(int x,int dad){
		fa[x]=dad;
		for(int v:e[x]) if(v!=dad) dfs(v,x);
	}
	int solve(int *p,int rt){
		dfs(rt,0);
		int s=0,tag[110]={};
		for(int i=1;i<=n;i++)
			if(i!=rt) tag[i]=fa[i]!=p[i],s+=tag[i];
		for(int i=1;i<=n;i++)
			if(i!=rt&&tag[fa[i]]&&!tag[i]) return 1e9;
		int d[110]={};
		vector<int> pre[110];
		for(int i=1;i<=n;i++){
			if(i!=rt&&tag[i]&&tag[fa[i]]){
				pre[fa[i]].push_back(i);
				d[i]++;
			}
			if(i!=rt&&tag[i]&&tag[p[i]]){
				pre[i].push_back(p[i]);
				d[p[i]]++;
			}
		}
		queue<int> q;int now=0;
		for(int i=1;i<=n;i++)
			if(i!=rt&&tag[i]&&!d[i]) q.push(i);
		while(!q.empty()){
			int u=q.front();q.pop();now++;
			for(int v:pre[u])
				if(--d[v]==0) q.push(v);
		}
		if(now==s) return s;
		else return 1e9;
	}
};
int main(){
	int T;read(T);
	while(T--){
		Tree T1,T2;
		read(n);
		T1.init();
		T2.init();
		int flag=1,ans=1e9;
		for(int i=1;i<=n;i++)
			if(T1.e[i]!=T2.e[i]) flag=0;
		if(flag){puts("0");continue;}
		for(int i=1;i<=n;i++){
			T2.dfs(i,0);
			if(T1.e[i].size()!=1) continue;
			for(int j=1;j<=n;j++)
				if(j!=i){
					Tree t=T1;
					t.e[*t.e[i].begin()].erase(i);
					t.e[i].clear();
					t.e[i].insert(j);
					t.e[j].insert(i);
					ans=min(ans,t.solve(T2.fa,i)+1);
				}
		}
		if(ans>1e8) ans=-1;
		cout<<ans<<'\n';
	}
	return 0;
}

```