#include<bits/stdc++.h>
#define cs const
#define pb push_back
using namespace std;
cs int N = 55;
cs int INF = 1e9 + 7;
int n, Ans; 
vector<int> E[N], G[N], tG[N];
void rebuild(int u, int v){
	for(int i=1; i<=n; i++)if(i!=u){
		G[i].clear();
		for(int e=0; e<(int)tG[i].size(); e++) 
		if(tG[i][e]!=u) G[i].pb(tG[i][e]);
	} if(u) G[u].clear(), G[u].pb(v), G[v].pb(u);
}
vector<int> T[N];
int A[N], B[N];
void fir_dfs(int u, int f){
	A[u]=f;
	for(int e=0,v;e<(int)E[u].size();e++) 
	if((v=E[u][e])!=f) fir_dfs(v,u);
} void sec_dfs(int u, int f){
	B[u]=f;
	for(int e=0,v;e<(int)G[u].size();e++)
	if((v=G[u][e])!=f) sec_dfs(v,u);
} bool nt[N]; int dg[N];
int topsort(){
	static int q[N]; int hd=1, tl=0;
	for(int i=1; i<=n; i++) 
	if(nt[i]&&!dg[i]) q[++tl]=i;
	while(hd<=tl){
		int x=q[hd++]; 
		for(int e=0,v; e<(int)T[x].size(); e++){
			v=T[x][e]; if(--dg[v]==0) q[++tl]=v;
		}
	} return tl;
}
int cope(int u){
	for(int i=1; i<=n; i++) T[i].clear(), dg[i]=0;
	fir_dfs(u,0); sec_dfs(u,0); int ct=0;
	for(int i=1; i<=n; i++) ct+=(nt[i]=(A[i]!=B[i]));
	if(!ct) return 0;
	for(int i=1; i<=n; i++){
		if(!nt[i]){ if(nt[A[i]]) return INF; continue; }
		if(nt[A[i]]) T[A[i]].pb(i),++dg[i];
		if(nt[B[i]]) T[i].pb(B[i]),++dg[B[i]];
	} if(topsort()!=ct) return INF; return ct;
}
void Main(){
	scanf("%d",&n); Ans=INF;
	for(int i=1; i<=n; i++)
	E[i].clear(), tG[i].clear();
	for(int i=1,u,v; i<n; i++)
	scanf("%d%d",&u,&v), E[u].pb(v), E[v].pb(u);
	for(int i=1,u,v; i<n; i++)
	scanf("%d%d",&u,&v), tG[u].pb(v), tG[v].pb(u);
	for(int u=1; u<=n; u++) 
	rebuild(0,0), Ans=min(Ans,cope(u));
	for(int u=1; u<=n; u++) 
	if(tG[u].size()==1){
		for(int v=1; v<=n; v++)
		rebuild(u,v), Ans=min(Ans,1+cope(u));
	} cout<<((Ans>=INF)?-1:Ans)<<'\n';
}
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	int T; scanf("%d",&T); 
	while(T--) Main(); return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 cs const
#define pb push_back
using namespace std;
cs int N = 55;
cs int INF = 1e9 + 7;
int n, Ans; 
vector<int> E[N], G[N], tG[N];
void rebuild(int u, int v){
	for(int i=1; i<=n; i++)if(i!=u){
		G[i].clear();
		for(int e=0; e<(int)tG[i].size(); e++) 
		if(tG[i][e]!=u) G[i].pb(tG[i][e]);
	} if(u) G[u].clear(), G[u].pb(v), G[v].pb(u);
}
vector<int> T[N];
int A[N], B[N];
void fir_dfs(int u, int f){
	A[u]=f;
	for(int e=0,v;e<(int)E[u].size();e++) 
	if((v=E[u][e])!=f) fir_dfs(v,u);
} void sec_dfs(int u, int f){
	B[u]=f;
	for(int e=0,v;e<(int)G[u].size();e++)
	if((v=G[u][e])!=f) sec_dfs(v,u);
} bool nt[N]; int dg[N];
int topsort(){
	static int q[N]; int hd=1, tl=0;
	for(int i=1; i<=n; i++) 
	if(nt[i]&&!dg[i]) q[++tl]=i;
	while(hd<=tl){
		int x=q[hd++]; 
		for(int e=0,v; e<(int)T[x].size(); e++){
			v=T[x][e]; if(--dg[v]==0) q[++tl]=v;
		}
	} return tl;
}
int cope(int u){
	for(int i=1; i<=n; i++) T[i].clear(), dg[i]=0;
	fir_dfs(u,0); sec_dfs(u,0); int ct=0;
	for(int i=1; i<=n; i++) ct+=(nt[i]=(A[i]!=B[i]));
	if(!ct) return 0;
	for(int i=1; i<=n; i++){
		if(!nt[i]){ if(nt[A[i]]) return INF; continue; }
		if(nt[A[i]]) T[A[i]].pb(i),++dg[i];
		if(nt[B[i]]) T[i].pb(B[i]),++dg[B[i]];
	} if(topsort()!=ct) return INF; return ct;
}
void Main(){
	scanf("%d",&n); Ans=INF;
	for(int i=1; i<=n; i++)
	E[i].clear(), tG[i].clear();
	for(int i=1,u,v; i<n; i++)
	scanf("%d%d",&u,&v), E[u].pb(v), E[v].pb(u);
	for(int i=1,u,v; i<n; i++)
	scanf("%d%d",&u,&v), tG[u].pb(v), tG[v].pb(u);
	for(int u=1; u<=n; u++) 
	rebuild(0,0), Ans=min(Ans,cope(u));
	for(int u=1; u<=n; u++) 
	if(tG[u].size()==1){
		for(int v=1; v<=n; v++)
		rebuild(u,v), Ans=min(Ans,1+cope(u));
	} cout<<((Ans>=INF)?-1:Ans)<<'\n';
}
int main(){
	#ifdef FSYolanda
	freopen("1.in","r",stdin);
	#endif
	int T; scanf("%d",&T); 
	while(T--) Main(); return 0;
}
```