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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
const int maxn = 60;

int T, n, ind[maxn], deg[maxn];
bool ban[maxn], vis[maxn];

vector<int> G[maxn], e1[maxn], e2[maxn];

void init()
{
	for(int i = 1; i <= n; ++i)
	{
		G[i].clear();
		ind[i] = 0;
		vis[i] = 0;
	}
}

void Clear()
{
	for(int i = 1; i <= n; ++i)
	{
		e1[i].clear();
		e2[i].clear();	
		deg[i] = 0;
	}
}

void dfs1(int x, int fa)
{
	int id;
	for(int i = 0; i < (int)e1[x].size(); ++i)
	{
		id = e1[x][i];
		if(id == fa || ban[id])	continue;
		if(!vis[id] && !vis[x])
		{
			G[id].push_back(x);
			++ ind[x];
		}
		dfs1(id, x);
	}
}

void dfs2(int x, int fa)
{
	int id;
	for(int i = 0; i < (int)e2[x].size(); ++i)
	{
		id = e2[x][i];
		if(id == fa)	continue;
		if(!vis[x] && !vis[id])
		{
			G[x].push_back(id);
			++ ind[id];
		}
		dfs2(id, x);
	}
}

int dfs3(int x, int fa)
{
	vis[x] = 1;
	int ret = 1;
	vector<int> :: iterator pt1 = e1[x].begin(), pt2 = e2[x].begin();
	while(pt1 != e1[x].end() && pt2 != e2[x].end())
	{
		if((*pt1) == fa)
		{
			++ pt1;
			continue;
		}
		if((*pt2) == fa)
		{
			++ pt2;
			continue;
		}
		if((*pt1) == (*pt2))
		{
			ret += dfs3((*pt1), x);
			++ pt1;
			++ pt2;
			continue;
		}
		if((*pt1) < (*pt2))
			++ pt1;
		else
			++ pt2;
	}
	return ret;
}

int que[maxn];

bool bfs()
{
	int h = 1, t = 0, s, id;
	for(int i = 1; i <= n; ++i)
		if(!ind[i])
		{
			que[++t] = i;
		}
	while(h <= t)
	{
		s = que[h++];
		for(int i = 0; i < (int)G[s].size(); ++i)
		{
			id = G[s][i];
			-- ind[id];
			if(!ind[id])
			{
				que[++t] = id;
			}
		}
	}
	return t == n;
}

int main()
{
	scanf("%d", &T);
	while(T --)
	{
		scanf("%d", &n);
		int u, v;
		for(int i = 1; i < n; ++i)
		{
			scanf("%d%d", &u, &v);
			e1[u].push_back(v);
			e1[v].push_back(u);
			++ deg[u];
			++ deg[v];
		}
		for(int i = 1; i < n; ++i)
		{
			scanf("%d%d", &u, &v);
			e2[u].push_back(v);
			e2[v].push_back(u);
		}
		for(int i = 1; i <= n; ++i)
		{
			sort(e1[i].begin(), e1[i].end());
			sort(e2[i].begin(), e2[i].end());
		}
		int ans = 100;
		for(int i = 1; i <= n; ++i)
		{
			init();
			int res = dfs3(i, 0);
			dfs1(i, 0);
			dfs2(i, 0);
			if(bfs())
			{
				ans = min(ans, n - res);
			}
		}
		for(int i = 1; i <= n; ++i)
			if(deg[i] == 1)
			{
				ban[i] = 1;
				for(int j = 1; j <= n; ++j)
					if(i != j)
					{
						init();
						int res = 0;
						for(int k = 0; k < (int)e2[i].size(); ++k)
							if(e2[i][k] == j)
							{
								res = dfs3(j, i);
								break;	
							}
						vis[i] = 1;
						dfs1(j, 0);
						dfs2(i, 0);
						if(bfs())
						{
							ans = min(ans, n - res);
						}
					}
				ban[i] = 0;
			}
		printf("%d\n", ans == 100? -1: ans);
		Clear();
	}
	return 0;
}