#include<bits/stdc++.h>
using namespace std;
#define y1 ___y1
const int N=105;
int ne[N],x1[N],T,y1[N],x2[N],y2[N],n,m,fi[N],tot,zz[N],d[N],Flag,flag[N],flag2[N];
struct Tree{
	int ne[N],tot,fi[N],zz[N],fa[N];
	void jb(int x,int y){
		ne[++tot]=fi[x];
		fi[x]=tot;
		zz[tot]=y;
		swap(x,y);
		ne[++tot]=fi[x];
		fi[x]=tot;
		zz[tot]=y;
	}
	void dfs(int x,int y){
		fa[x]=y;
		for (int i=fi[x];i;i=ne[i])
			if (zz[i]!=y)dfs(zz[i],x);
	}
}T1,T2;
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
void dfs(int x){
	flag[x]=1;
	for (int i=fi[x];i;i=ne[i]){
		if (flag[zz[i]]==1)Flag=1;
		if (!flag[zz[i]])dfs(zz[i]);
	}
	flag[x]=2;
}
int main(){
	scanf("%d",&T);
	while (T--){
		scanf("%d",&n);
		memset(T1.fi,0,sizeof T1.fi);
		memset(T2.fi,0,sizeof T2.fi);
		T1.tot=T2.tot=0;
		memset(d,0,sizeof d);
		for (int i=1;i<n;i++)scanf("%d%d",&x1[i],&y1[i]),d[x1[i]]++,d[y1[i]]++;
		for (int i=1;i<n;i++)scanf("%d%d",&x2[i],&y2[i]);
		for (int i=1;i<n;i++)T1.jb(x1[i],y1[i]),T2.jb(x2[i],y2[i]);
		T1.dfs(1,0);T2.dfs(1,0);
		int fflag=0;
		for (int i=1;i<=n;i++)fflag|=T1.fa[i]!=T2.fa[i];
		if (!fflag){
			puts("0");
			continue;
		}
		int ans=1e9;
		for (int i=1;i<=n;i++)
			if (d[i]==1)for (int j=1;j<=n;j++)
				if (j!=i){
					T1.tot=0;
					memset(T1.fi,0,sizeof T1.fi);
					for (int k=1;k<n;k++)
						if (x1[k]!=i&&y1[k]!=i)T1.jb(x1[k],y1[k]);
					T1.jb(i,j);
					memset(fi,0,sizeof fi);
					tot=0;
					memset(flag,0,sizeof flag);
					T1.dfs(i,0);
					T2.dfs(i,0);
					Flag=0;int cnt=0;
					for (int k=1;k<=n;k++)flag2[k]=T1.fa[k]!=T2.fa[k];
					for (int k=1;k<=n;k++)cnt+=flag2[k];
					for (int k=1;k<=n;k++){
						if (!flag2[k]&&flag2[T1.fa[k]])Flag=1;
						if (!flag2[k]&&flag2[T2.fa[k]])Flag=1;
						if (flag2[k]&&flag2[T1.fa[k]])jb(k,T1.fa[k]);
						if (flag2[k]&&flag2[T2.fa[k]])jb(T2.fa[k],k);
					}
					for (int k=1;k<=n;k++)
						if (!flag[k])dfs(k);
					if (!Flag)ans=min(ans,cnt);
				}
		if (ans==1e9)puts("-1");
		else printf("%d\n",ans+1);
	}
	return 0;
}

### Prompt

Generate 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>
using namespace std;
#define y1 ___y1
const int N=105;
int ne[N],x1[N],T,y1[N],x2[N],y2[N],n,m,fi[N],tot,zz[N],d[N],Flag,flag[N],flag2[N];
struct Tree{
	int ne[N],tot,fi[N],zz[N],fa[N];
	void jb(int x,int y){
		ne[++tot]=fi[x];
		fi[x]=tot;
		zz[tot]=y;
		swap(x,y);
		ne[++tot]=fi[x];
		fi[x]=tot;
		zz[tot]=y;
	}
	void dfs(int x,int y){
		fa[x]=y;
		for (int i=fi[x];i;i=ne[i])
			if (zz[i]!=y)dfs(zz[i],x);
	}
}T1,T2;
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
void dfs(int x){
	flag[x]=1;
	for (int i=fi[x];i;i=ne[i]){
		if (flag[zz[i]]==1)Flag=1;
		if (!flag[zz[i]])dfs(zz[i]);
	}
	flag[x]=2;
}
int main(){
	scanf("%d",&T);
	while (T--){
		scanf("%d",&n);
		memset(T1.fi,0,sizeof T1.fi);
		memset(T2.fi,0,sizeof T2.fi);
		T1.tot=T2.tot=0;
		memset(d,0,sizeof d);
		for (int i=1;i<n;i++)scanf("%d%d",&x1[i],&y1[i]),d[x1[i]]++,d[y1[i]]++;
		for (int i=1;i<n;i++)scanf("%d%d",&x2[i],&y2[i]);
		for (int i=1;i<n;i++)T1.jb(x1[i],y1[i]),T2.jb(x2[i],y2[i]);
		T1.dfs(1,0);T2.dfs(1,0);
		int fflag=0;
		for (int i=1;i<=n;i++)fflag|=T1.fa[i]!=T2.fa[i];
		if (!fflag){
			puts("0");
			continue;
		}
		int ans=1e9;
		for (int i=1;i<=n;i++)
			if (d[i]==1)for (int j=1;j<=n;j++)
				if (j!=i){
					T1.tot=0;
					memset(T1.fi,0,sizeof T1.fi);
					for (int k=1;k<n;k++)
						if (x1[k]!=i&&y1[k]!=i)T1.jb(x1[k],y1[k]);
					T1.jb(i,j);
					memset(fi,0,sizeof fi);
					tot=0;
					memset(flag,0,sizeof flag);
					T1.dfs(i,0);
					T2.dfs(i,0);
					Flag=0;int cnt=0;
					for (int k=1;k<=n;k++)flag2[k]=T1.fa[k]!=T2.fa[k];
					for (int k=1;k<=n;k++)cnt+=flag2[k];
					for (int k=1;k<=n;k++){
						if (!flag2[k]&&flag2[T1.fa[k]])Flag=1;
						if (!flag2[k]&&flag2[T2.fa[k]])Flag=1;
						if (flag2[k]&&flag2[T1.fa[k]])jb(k,T1.fa[k]);
						if (flag2[k]&&flag2[T2.fa[k]])jb(T2.fa[k],k);
					}
					for (int k=1;k<=n;k++)
						if (!flag[k])dfs(k);
					if (!Flag)ans=min(ans,cnt);
				}
		if (ans==1e9)puts("-1");
		else printf("%d\n",ans+1);
	}
	return 0;
}

```