#include <bits/stdc++.h>
#define rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define foreach(c,itr) for (__typeof((c).begin()) itr=(c).begin();itr!=(c).end();++itr)
using namespace std;
int t,n,m,i,j,fa1[55],fa2[55],deg[55],ans,isl[55];
vector<int> ga[55],gb[55];
void dfs1(int x,int fa){
	fa1[x]=fa;
	foreach(ga[x],i){
		if(*i!=fa) dfs1(*i,x);
	}
}
void dfs2(int x,int fa){
	fa2[x]=fa;
	foreach(gb[x],i){
		if(*i!=fa) dfs2(*i,x);
	}
}
void solve(int x,int y){
	dfs1(y,0);dfs2(x,0);
	fa1[y]=x;fa1[x]=0;
	memset(deg,0,sizeof(deg));
	int i,j,c=0;
	rep(i,n) if(i!=x) deg[fa1[i]]++;
	rep(i,n) if(i!=x&&fa1[i]!=fa2[i]) c++;
	rep(i,c){
		rep(j,n){
			if(j==x) continue;
			if(fa1[j]==fa2[j]) continue;
			if(deg[j]) continue;
			if(fa1[fa2[j]]!=fa2[fa2[j]]) continue;
			break;
		}
		if(j>n) return;
		deg[fa1[j]]--;
		deg[fa2[j]]++;
		fa1[j]=fa2[j];
	}
	ans=min(ans,c+1);
}
int main(){
//	freopen("1.in","r",stdin); 
	cin>>t;
	while(t--){
		cin>>n;
		rep(i,n){
			isl[i]=0;
			ga[i].clear();
			gb[i].clear();
		}
		rep(i,n-1){
			int x,y;
			cin>>x>>y;
			ga[x].push_back(y);
			ga[y].push_back(x);
			isl[x]++;isl[y]++;
		}
		rep(i,n-1){
			int x,y;
			cin>>x>>y;
			gb[x].push_back(y);
			gb[y].push_back(x);
		}
		int f=1;
		dfs1(1,0);dfs2(1,0);
		rep(i,n)if(fa1[i]!=fa2[i]) f=0;
		if(f){
			puts("0");
			continue; 
		}
		ans=0x3f3f3f3f;
		rep(i,n)if(isl[i]==1)rep(j,n)if(i!=j){
			solve(i,j);
		}
		if(ans==0x3f3f3f3f){
			puts("-1");
		}
		else{
			printf("%d\n",ans);
		}
	}
	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 rep(i,n) for ((i)=1;(i)<=(n);(i)++)
#define foreach(c,itr) for (__typeof((c).begin()) itr=(c).begin();itr!=(c).end();++itr)
using namespace std;
int t,n,m,i,j,fa1[55],fa2[55],deg[55],ans,isl[55];
vector<int> ga[55],gb[55];
void dfs1(int x,int fa){
	fa1[x]=fa;
	foreach(ga[x],i){
		if(*i!=fa) dfs1(*i,x);
	}
}
void dfs2(int x,int fa){
	fa2[x]=fa;
	foreach(gb[x],i){
		if(*i!=fa) dfs2(*i,x);
	}
}
void solve(int x,int y){
	dfs1(y,0);dfs2(x,0);
	fa1[y]=x;fa1[x]=0;
	memset(deg,0,sizeof(deg));
	int i,j,c=0;
	rep(i,n) if(i!=x) deg[fa1[i]]++;
	rep(i,n) if(i!=x&&fa1[i]!=fa2[i]) c++;
	rep(i,c){
		rep(j,n){
			if(j==x) continue;
			if(fa1[j]==fa2[j]) continue;
			if(deg[j]) continue;
			if(fa1[fa2[j]]!=fa2[fa2[j]]) continue;
			break;
		}
		if(j>n) return;
		deg[fa1[j]]--;
		deg[fa2[j]]++;
		fa1[j]=fa2[j];
	}
	ans=min(ans,c+1);
}
int main(){
//	freopen("1.in","r",stdin); 
	cin>>t;
	while(t--){
		cin>>n;
		rep(i,n){
			isl[i]=0;
			ga[i].clear();
			gb[i].clear();
		}
		rep(i,n-1){
			int x,y;
			cin>>x>>y;
			ga[x].push_back(y);
			ga[y].push_back(x);
			isl[x]++;isl[y]++;
		}
		rep(i,n-1){
			int x,y;
			cin>>x>>y;
			gb[x].push_back(y);
			gb[y].push_back(x);
		}
		int f=1;
		dfs1(1,0);dfs2(1,0);
		rep(i,n)if(fa1[i]!=fa2[i]) f=0;
		if(f){
			puts("0");
			continue; 
		}
		ans=0x3f3f3f3f;
		rep(i,n)if(isl[i]==1)rep(j,n)if(i!=j){
			solve(i,j);
		}
		if(ans==0x3f3f3f3f){
			puts("-1");
		}
		else{
			printf("%d\n",ans);
		}
	}
	return 0;
}
```