#include<bits/stdc++.h>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
using namespace std;
typedef long long ll;
typedef vector<int> vt;
int i,j,T,d[55],fa[55],n,x,y,ans,cost;
vt a[55],b[55];
bool bz[55],w[55][55];
void dfs(int x,int y){
	fa[x]=y;
	if (bz[y]&&w[x][y]) bz[x]=1;
	for(int i=0;i<b[x].size();i++){
		int g=b[x][i];
		if (g!=y) dfs(g,x);
	}
}
void work(){
	memset(w,0,sizeof(w));
	scanf("%d",&n);
	fo(i,1,n) a[i].clear(),b[i].clear();
	fo(i,1,n-1){
		scanf("%d%d",&x,&y);
		w[x][y]=w[y][x]=1;
		a[x].push_back(y),a[y].push_back(x);
	}
	fo(i,1,n-1){
		scanf("%d%d",&x,&y);
		b[x].push_back(y),b[y].push_back(x);
	}
	ans=n+1;
	fo(i,1,n){
		memset(bz,0,sizeof(bz));
		bz[i]=1;
		dfs(i,0);
		fo(x,1,n) d[x]=a[x].size();
		cost=0;
		for(;;){
			fo(x,1,n) if (!bz[x]&&d[x]==1&&bz[fa[x]]) break;
			if (x>n) break;
			bz[x]=1,cost++;
			for(j=0;j<a[x].size();j++) d[a[x][j]]--;
 		}
		fo(x,1,n) if (!bz[x]) break;
		if (x>n) ans=min(ans,cost);
		if (a[i].size()>1||x>n||ans<=n) continue;
		memset(bz,0,sizeof(bz));
		bz[0]=1;
		fo(x,1,n) d[x]=a[x].size();
		for(;;){
			fo(x,1,n) if (!bz[x]&&d[x]<=1&&bz[fa[x]]) break;
			if (x>n) break;
			bz[x]=1;
			for(j=0;j<a[x].size();j++) d[a[x][j]]--;
 		}
 		fo(x,1,n) if (!bz[x]) break;
		if (x>n) ans=n;
	}
	printf("%d\n",ans>n?-1:ans);
}
int main(){
	for(scanf("%d",&T);T--;) work();
}

### 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<bits/stdc++.h>
#define fo(i,a,b) for(i=a;i<=b;i++)
#define fd(i,a,b) for(i=a;i>=b;i--)
#define min(a,b) (a<b?a:b)
using namespace std;
typedef long long ll;
typedef vector<int> vt;
int i,j,T,d[55],fa[55],n,x,y,ans,cost;
vt a[55],b[55];
bool bz[55],w[55][55];
void dfs(int x,int y){
	fa[x]=y;
	if (bz[y]&&w[x][y]) bz[x]=1;
	for(int i=0;i<b[x].size();i++){
		int g=b[x][i];
		if (g!=y) dfs(g,x);
	}
}
void work(){
	memset(w,0,sizeof(w));
	scanf("%d",&n);
	fo(i,1,n) a[i].clear(),b[i].clear();
	fo(i,1,n-1){
		scanf("%d%d",&x,&y);
		w[x][y]=w[y][x]=1;
		a[x].push_back(y),a[y].push_back(x);
	}
	fo(i,1,n-1){
		scanf("%d%d",&x,&y);
		b[x].push_back(y),b[y].push_back(x);
	}
	ans=n+1;
	fo(i,1,n){
		memset(bz,0,sizeof(bz));
		bz[i]=1;
		dfs(i,0);
		fo(x,1,n) d[x]=a[x].size();
		cost=0;
		for(;;){
			fo(x,1,n) if (!bz[x]&&d[x]==1&&bz[fa[x]]) break;
			if (x>n) break;
			bz[x]=1,cost++;
			for(j=0;j<a[x].size();j++) d[a[x][j]]--;
 		}
		fo(x,1,n) if (!bz[x]) break;
		if (x>n) ans=min(ans,cost);
		if (a[i].size()>1||x>n||ans<=n) continue;
		memset(bz,0,sizeof(bz));
		bz[0]=1;
		fo(x,1,n) d[x]=a[x].size();
		for(;;){
			fo(x,1,n) if (!bz[x]&&d[x]<=1&&bz[fa[x]]) break;
			if (x>n) break;
			bz[x]=1;
			for(j=0;j<a[x].size();j++) d[a[x][j]]--;
 		}
 		fo(x,1,n) if (!bz[x]) break;
		if (x>n) ans=n;
	}
	printf("%d\n",ans>n?-1:ans);
}
int main(){
	for(scanf("%d",&T);T--;) work();
}

```