#include<cstdio>
const int MaxN=55;
int N;
struct Edge{int to;Edge*next;};
struct Tree{
	Edge E[MaxN*2],*ne,*fir[MaxN];
	void link(int u,int v){*ne=(Edge){v,fir[u]};fir[u]=ne++;}
	void init(int n){
		ne=E;for(int i=1;i<=N;i++)fir[i]=0;
	}
	void read(int n){
		init(n);
		for(int i=1,u,v;i<N;i++)scanf("%d%d",&u,&v),link(u,v),link(v,u);
	}
}A,B,I;
bool vis[MaxN],del[MaxN],in[MaxN];
int deg[MaxN],stk[MaxN];
void ff(int i){
	if(vis[i])return;
	vis[i]=del[i]=1;
	for(Edge*e=I.fir[i];e;e=e->next)ff(e->to);
}
int main(){
	int T;scanf("%d",&T);
	while(T--){
		scanf("%d",&N);
		A.read(N);
		B.read(N);
		I.init(N);
		for(int i=1;i<=N;i++)vis[i]=0;
		for(int i=1;i<=N;i++){
			for(Edge*e=A.fir[i];e;e=e->next)vis[e->to]=1;
			for(Edge*e=B.fir[i];e;e=e->next)if(vis[e->to])I.link(i,e->to);
			for(Edge*e=A.fir[i];e;e=e->next)vis[e->to]=0;
		}
		int ans=-1;
		for(int ty=0;ty<2;ty++){
			for(int i=1;i<=N;i++)if(ty?!A.fir[i]->next:!vis[i]){
				for(int j=1;j<=N;j++){
					del[j]=deg[j]=in[j]=0;
					for(Edge*e=A.fir[j];e;e=e->next)deg[j]++;
				}
				//printf("(");
				//for(int j=1;j<=N;j++)if(del[j])printf(" %d ",j);puts(")");
				int*top=stk,tot=0;
				if(ty)del[*top++=i]=1;
				else{
					ff(i);
					for(int j=1;j<=N;j++)if(del[j])
						for(Edge*e=B.fir[j];e;e=e->next)if(!del[e->to])in[e->to]=1,deg[e->to]<2?del[*top++=e->to]=1:1;
				}
				while(top>stk){
					int x=*--top;
				//	if(ty)printf("i=%d del %d\n",i,x);
					for(Edge*e=A.fir[x];e;e=e->next)if(!del[e->to]&&--deg[e->to]<2&&in[e->to])del[*top++=e->to]=1;
					for(Edge*e=B.fir[x];e;e=e->next)if(!del[e->to]&&(in[e->to]=1)&&deg[e->to]<2)del[*top++=e->to]=1;
					tot++;
				}
				for(int j=1;j<=N;j++)if(!del[j])tot=-1;
				if(tot>-1&&(ans==-1||tot<ans))ans=tot;
			}
		}
		printf("%d\n",ans);
	}
}
### Prompt

Please formulate 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<cstdio>
const int MaxN=55;
int N;
struct Edge{int to;Edge*next;};
struct Tree{
	Edge E[MaxN*2],*ne,*fir[MaxN];
	void link(int u,int v){*ne=(Edge){v,fir[u]};fir[u]=ne++;}
	void init(int n){
		ne=E;for(int i=1;i<=N;i++)fir[i]=0;
	}
	void read(int n){
		init(n);
		for(int i=1,u,v;i<N;i++)scanf("%d%d",&u,&v),link(u,v),link(v,u);
	}
}A,B,I;
bool vis[MaxN],del[MaxN],in[MaxN];
int deg[MaxN],stk[MaxN];
void ff(int i){
	if(vis[i])return;
	vis[i]=del[i]=1;
	for(Edge*e=I.fir[i];e;e=e->next)ff(e->to);
}
int main(){
	int T;scanf("%d",&T);
	while(T--){
		scanf("%d",&N);
		A.read(N);
		B.read(N);
		I.init(N);
		for(int i=1;i<=N;i++)vis[i]=0;
		for(int i=1;i<=N;i++){
			for(Edge*e=A.fir[i];e;e=e->next)vis[e->to]=1;
			for(Edge*e=B.fir[i];e;e=e->next)if(vis[e->to])I.link(i,e->to);
			for(Edge*e=A.fir[i];e;e=e->next)vis[e->to]=0;
		}
		int ans=-1;
		for(int ty=0;ty<2;ty++){
			for(int i=1;i<=N;i++)if(ty?!A.fir[i]->next:!vis[i]){
				for(int j=1;j<=N;j++){
					del[j]=deg[j]=in[j]=0;
					for(Edge*e=A.fir[j];e;e=e->next)deg[j]++;
				}
				//printf("(");
				//for(int j=1;j<=N;j++)if(del[j])printf(" %d ",j);puts(")");
				int*top=stk,tot=0;
				if(ty)del[*top++=i]=1;
				else{
					ff(i);
					for(int j=1;j<=N;j++)if(del[j])
						for(Edge*e=B.fir[j];e;e=e->next)if(!del[e->to])in[e->to]=1,deg[e->to]<2?del[*top++=e->to]=1:1;
				}
				while(top>stk){
					int x=*--top;
				//	if(ty)printf("i=%d del %d\n",i,x);
					for(Edge*e=A.fir[x];e;e=e->next)if(!del[e->to]&&--deg[e->to]<2&&in[e->to])del[*top++=e->to]=1;
					for(Edge*e=B.fir[x];e;e=e->next)if(!del[e->to]&&(in[e->to]=1)&&deg[e->to]<2)del[*top++=e->to]=1;
					tot++;
				}
				for(int j=1;j<=N;j++)if(!del[j])tot=-1;
				if(tot>-1&&(ans==-1||tot<ans))ans=tot;
			}
		}
		printf("%d\n",ans);
	}
}
```