#include<bits/stdc++.h>
#define LL long long
#define db double

using namespace std;
const int N=55;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
struct graph
{
	int to[N<<2],nt[N<<2],ban[N<<2],hd[N],dg[N],tot;
	graph(){tot=1;}
	void adde(int x,int y){++tot,to[tot]=y,nt[tot]=hd[x],hd[x]=tot,++dg[y];}
	void wd()
	{
		int x=to[tot],y=to[tot-1];
		hd[x]=nt[hd[x]],hd[y]=nt[hd[y]],--dg[x],--dg[y];
		tot-=2;
	}
}t0,t1,g0;
bool z2[N][N],mk[N],ban[N];
int n,ans,na,fa[N],f2[N];
queue<int> q;
void dfs1(int x)
{
	for(int i=t0.hd[x];i;i=t0.nt[i])
	{
		int y=t0.to[i];
		if(t0.ban[i>>1]||y==fa[x]) continue;
		fa[y]=x,mk[y]=mk[x]&z2[x][y],dfs1(y);
	}
}
void dfs2(int x)
{
	for(int i=t1.hd[x];i;i=t1.nt[i])
	{
		int y=t1.to[i];
		if(t1.ban[i>>1]||y==f2[x]) continue;
		f2[y]=x,dfs2(y);
	}
}
void wk(int i,int o)
{
	memset(mk,0,sizeof(bool)*(n+1));
	fa[i]=f2[i]=0,mk[0]=mk[i]=1;
	dfs1(i),dfs2(i),mk[i]=o;
	memset(g0.hd,0,sizeof(int)*(n+1)),memset(g0.dg,0,sizeof(int)*(n+1)),g0.tot=1;
	na=n;
	for(int j=1;j<=n;++j)
	{
		na-=mk[j];
		if(fa[j]!=i&&!mk[j]&&!mk[fa[j]]) g0.adde(fa[j],j);
		if(f2[j]!=i&&!mk[j]&&!mk[f2[j]]) g0.adde(j,f2[j]);
	}
	for(int j=1;j<=n;++j)
		if(!g0.dg[j]) q.push(j);
	int rs=n;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		--rs;
		for(int k=g0.hd[x];k;k=g0.nt[k])
		{
			int y=g0.to[k];
			--g0.dg[y];
			if(!g0.dg[y]) q.push(y);
		}
	}
	if(!rs) ans=min(ans,na);
}

int main()
{
//// /bb
	int T=rd();
	while(T--)
	{
		n=rd();
		memset(t0.hd,0,sizeof(int)*(n+1)),memset(t0.dg,0,sizeof(int)*(n+1)),t0.tot=1;
		for(int i=1;i<n;++i)
		{
			int x=rd(),y=rd();
			t0.adde(x,y),t0.adde(y,x);
		}
		memset(t1.hd,0,sizeof(int)*(n+1)),t1.tot=1;
		memset(z2,0,sizeof(z2));
		for(int i=1;i<n;++i)
		{
			int x=rd(),y=rd();
			t1.adde(x,y),t1.adde(y,x),z2[x][y]=z2[y][x]=1;
		}
		ans=n+1;
		for(int i=1;i<=n;++i) wk(i,1);
		for(int x=1;x<=n;++x)
		{
			if(t0.dg[x]>1) continue;
			int j=t0.hd[x];
			t0.ban[j>>1]=1;
			for(int y=1;y<=n;++y)
				if(x!=y) t0.adde(x,y),t0.adde(y,x),wk(x,0),t0.wd();
			t0.ban[j>>1]=0;
		}
		printf("%d\n",ans<=n?ans:-1);
	}
	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 LL long long
#define db double

using namespace std;
const int N=55;
int rd()
{
    int x=0,w=1;char ch=0;
    while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+(ch^48);ch=getchar();}
    return x*w;
}
struct graph
{
	int to[N<<2],nt[N<<2],ban[N<<2],hd[N],dg[N],tot;
	graph(){tot=1;}
	void adde(int x,int y){++tot,to[tot]=y,nt[tot]=hd[x],hd[x]=tot,++dg[y];}
	void wd()
	{
		int x=to[tot],y=to[tot-1];
		hd[x]=nt[hd[x]],hd[y]=nt[hd[y]],--dg[x],--dg[y];
		tot-=2;
	}
}t0,t1,g0;
bool z2[N][N],mk[N],ban[N];
int n,ans,na,fa[N],f2[N];
queue<int> q;
void dfs1(int x)
{
	for(int i=t0.hd[x];i;i=t0.nt[i])
	{
		int y=t0.to[i];
		if(t0.ban[i>>1]||y==fa[x]) continue;
		fa[y]=x,mk[y]=mk[x]&z2[x][y],dfs1(y);
	}
}
void dfs2(int x)
{
	for(int i=t1.hd[x];i;i=t1.nt[i])
	{
		int y=t1.to[i];
		if(t1.ban[i>>1]||y==f2[x]) continue;
		f2[y]=x,dfs2(y);
	}
}
void wk(int i,int o)
{
	memset(mk,0,sizeof(bool)*(n+1));
	fa[i]=f2[i]=0,mk[0]=mk[i]=1;
	dfs1(i),dfs2(i),mk[i]=o;
	memset(g0.hd,0,sizeof(int)*(n+1)),memset(g0.dg,0,sizeof(int)*(n+1)),g0.tot=1;
	na=n;
	for(int j=1;j<=n;++j)
	{
		na-=mk[j];
		if(fa[j]!=i&&!mk[j]&&!mk[fa[j]]) g0.adde(fa[j],j);
		if(f2[j]!=i&&!mk[j]&&!mk[f2[j]]) g0.adde(j,f2[j]);
	}
	for(int j=1;j<=n;++j)
		if(!g0.dg[j]) q.push(j);
	int rs=n;
	while(!q.empty())
	{
		int x=q.front();
		q.pop();
		--rs;
		for(int k=g0.hd[x];k;k=g0.nt[k])
		{
			int y=g0.to[k];
			--g0.dg[y];
			if(!g0.dg[y]) q.push(y);
		}
	}
	if(!rs) ans=min(ans,na);
}

int main()
{
//// /bb
	int T=rd();
	while(T--)
	{
		n=rd();
		memset(t0.hd,0,sizeof(int)*(n+1)),memset(t0.dg,0,sizeof(int)*(n+1)),t0.tot=1;
		for(int i=1;i<n;++i)
		{
			int x=rd(),y=rd();
			t0.adde(x,y),t0.adde(y,x);
		}
		memset(t1.hd,0,sizeof(int)*(n+1)),t1.tot=1;
		memset(z2,0,sizeof(z2));
		for(int i=1;i<n;++i)
		{
			int x=rd(),y=rd();
			t1.adde(x,y),t1.adde(y,x),z2[x][y]=z2[y][x]=1;
		}
		ans=n+1;
		for(int i=1;i<=n;++i) wk(i,1);
		for(int x=1;x<=n;++x)
		{
			if(t0.dg[x]>1) continue;
			int j=t0.hd[x];
			t0.ban[j>>1]=1;
			for(int y=1;y<=n;++y)
				if(x!=y) t0.adde(x,y),t0.adde(y,x),wk(x,0),t0.wd();
			t0.ban[j>>1]=0;
		}
		printf("%d\n",ans<=n?ans:-1);
	}
	return 0;
}

```